Question 14 Marks
In Fig. $\text{AB}=\text{AC}$ and $\angle\text{ACD}=105^\circ,$ find $\angle\text{BAC}.$
AnswerWe have,$\text{AB}=\text{AC}$ and $\angle\text{ACD}=(10)5^\circ$
Since, $\angle\text{BCD}=180^\circ=\text{Straight angle}$$\angle\text{BCA}+\angle\text{ACD}=180^\circ$
$\angle\text{BCA}+(10)5^\circ=180^\circ$
$\angle\text{BCA}=180^\circ-(10)5^\circ$
$\angle\text{BCA}=75^\circ$
And also,$\triangle\text{ABC}$ is an isosceles triangle [AB = AC]
$\angle\text{ABC}=\angle\text{ACB}$ [Angles opposite to equal sides are equal]
From (i), we have$\angle\text{ACB}=75^\circ$
$\angle\text{ABC}=\angle\text{ACB}=75^\circ$
And also, Sum of Interior angles of a triangle = 180°$\angle\text{ABC}=\angle\text{BCA}+\angle\text{CAB}=180^\circ$
$75^\circ+75^\circ+\angle\text{CAB}=180^\circ$
$150^\circ+\angle\text{BAC}=180^\circ$
$\angle\text{BAC}=180^\circ-150^\circ=30^\circ$
$\angle\text{BAC}=30^\circ$
View full question & answer→Question 24 Marks
Prove that in a quadrilateral the sum of all the sides is geater than the sum of its diagonals.
Answer
Diagonal AC and BD is joined. Since sum of any two sides of a triangles is greater than the thirdside. Therefore, In $\triangle\text{ABC}$$\text{AB}+\text{BC}>\text{AC}\ ....(1)$
In $\triangle\text{ACD}$$\text{AD}+\text{DC}>\text{AC}\ .....(2)$
In $\triangle\text{ABD}$$\text{AB}+\text{AD}>\text{BD}\ .....(3)$
andin $\triangle\text{BCD}$$\text{BC}+\text{CD}>\text{BD}\ ......(4)$
adding (1), (2), (3) and (4) we get$2\text{AD}+2\text{DC}+2 \text{AB}+2\text{BC}>2\text{AC}+2\text{BD}$
$\Rightarrow\text{AB}+\text{BC}+\text{CD}+\text{DA}>\text{AC}+\text{BD}$
Hence proved. View full question & answer→Question 34 Marks
In a $\triangle\text{ABC},$ if $\text{AB}=\text{AC}$ and $\angle\text{B}=70^\circ.$ Find $\angle\text{A}.$
AnswerIn a $\triangle\text{ABC},$ if $\text{AB}=\text{AC}$ and $\angle\text{B}=70^\circ$ Since, $\text{AB}=\text{AC }\triangle\text{ABC}$ is an isosceles triangle$\angle\text{B}=\angle\text{C}$ [Angles opposite to equal sides are equal]
$\angle\text{B}=\angle\text{C}=70^\circ$
And also, Sum of angles in a triangle = 180°$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{A}+70^\circ+70^\circ=180^\circ$
$\angle\text{A}=180^\circ-140^\circ$
$\angle\text{A}=40^\circ$
View full question & answer→Question 44 Marks
ABC is a triangle in which BE and CF are, respectively, the perpendiculals to the side AC and AB. if $\text{BE}=\text{CF},$ prove that $\triangle\text{ABC}$ is isosceles.
Answer
In $\triangle\text{BEC}$ and $\triangle\text{CFB}$$\text{BC}=\text{BC}$ [commonj hypotenuse]
$\angle\text{BFC}=\angle\text{CEB}=90^\circ$ [given]
$\text{BE}=\text{CF}$ [given]
By RHS congurence criterion $\triangle\text{BEC}\cong\triangle\text{CFB}$$\therefore\angle\text{B}=\angle\text{C}$ [c.p.c.t]
$\Rightarrow\text{AB}=\text{AC}$
$\Rightarrow\triangle\text{ABC}$ is isosceles. View full question & answer→Question 54 Marks
ABC is a triangle in which $\angle\text{B}=2\angle\text{C}.$ D is a point on BC such that AD bisects $\angle\text{BAC}$ and $\text{AB}=\text{CD}.$ Prove that $\angle\text{BAC}=72^\circ.$
AnswerGiven that in ABC, $\angle\text{B}=2\ \angle\text{C}$ and D is a point on BC such that AD bisectors $\angle\text{BAC}$ and $\text{AB}=\text{CD}.$ 
We have to prove that $\angle\text{BAC}=72^\circ$ Now, draw the angular bisector of $\angle\text{ABC},$ which meets AC in P. Join PD Let $\text{C}=\angle\text{ACB}=\text{y}$$\angle\text{B}=\angle\text{ABC}=2\angle\text{C}=2\text{y}$ and also
Let $\angle\text{BAD}=\angle\text{DAC}$$\angle\text{BAC}=2\text{x}$ $[$ AD is the bisector of $\angle\text{BAC}]$
Now, in $\triangle\text{BPC},$ $\angle\text{CBP}=\text{y}$ $[$ BP is the bisector of $\angle\text{ABC}]$
$\angle\text{PCB}=\text{y}$
$\angle\text{CBP}=\angle\text{PCB}=\text{y}[\text{PC}=\text{BP]}$
Consider, $\triangle\text{ABP}$ and $\triangle\text{DCP},$ we have$\triangle\text{ABP}=\triangle\text{DCP}=\text{y}$
$\text{AB}=\text{DC}$ [Given]
And $\text{PC}=\text{BP}$ [From above] So, by SAS congruence criterion, we have $\triangle\text{ABP}\cong\triangle\text{DCP}$ Now,$\angle\text{BAP}=\angle\text{CDF}$ and $\text{AP}=\text{DP}$ [Corresponding parts of congruent triangles are equal]
$\angle\text{BAP}=\angle\text{CDP}=2$
Consider, $\triangle\text{APD},$ We have $\text{AP}=\text{DP}$$\angle\text{ADP}=\angle\text{DAP}$
But $\angle\text{DAP}=\text{x}$$\angle\text{ADP}=\angle\text{DAP}=\text{x}$
Now, In $\triangle\text{ABD}.$$\angle\text{ABD}+\angle\text{BAD}+\angle\text{ADB}=180^\circ$
And also $\angle\text{ADB}+\angle\text{ADC}=180^\circ$ [Straight angle] From the above two equations, we get$\angle\text{ABD}+\angle\text{BAD}+\angle\text{ADB}+\angle\text{ADC}$
$2\text{y}+\text{x}=\angle\text{ADP}+\angle\text{PDC}$
$2\text{y}+\text{x}=\text{x}+2\text{y}$
$2\text{y}=2\text{x}$
$\text{y}=\text{x}\text{ (or})\text{ x}=\text{y}$
We know, Sum of angles in a triangle = 180° So, In $\triangle\text{ABC},$$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$2\text{x}+2\text{y}+\text{y}=180^\circ$ $[\angle\text{A}=2\text{x},\angle\text{B}=2\text{y},\angle\text{C}=\text{y}]$
$2(\text{y}+3\text{y}=180^\circ[\text{x}=\text{y}]$
$5\text{y}=180^\circ$
$\text{y}=36^\circ$
Now, $\angle\text{A}=\angle\text{BAC}=2\times36^\circ=72^\circ$ View full question & answer→Question 64 Marks
In Fig. prove that:
AnswerTo prove,
- CD + DA + AB + BC > 2AC
- CD + DA + AB > BC
From the given figure,
We know that, in a triangle sum of any two sides is greater than the third side.
- So,
In $\triangle\text{ABC},$ we have
AB + BC > AC ....(i)
In $\triangle\text{ADC},$ we have
CD + DA > AC ....(ii)
Adding (i) and (ii), we get
AB + BC + CD + DA > AC + AC
AB + BC + CD + DA > 2AC
- Now, In $\triangle\text{ABC},$ we have,
AB + AC > BC ...(iii)
And in $\triangle\text{ADC},$ we have
CD + DA > AC
Add AB on both sides
CD + DA + AB > AC + AB
From equation (iii) and (iv), we get,
CD + DA + AB > AC + AB > BC
CD + DA + AB > BC
Hence proved. View full question & answer→Question 74 Marks
ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.
AnswerIn the triangle ABC it is given that $\text{AB}=\text{AC},$ BE and CF are medians.
We have to show that $\text{BE}=\text{CF}$ To show $\text{BE}=\text{CF}$ we will show that $\triangle\text{BFC}\cong\triangle\text{BEC}$ In triangle $\triangle\text{BFC}$ and $\triangle\text{BEC}$ As $\text{AB}=\text{AC},$ so$\angle\text{FBC}=\angle\text{ECF}\ .....(1)$
$\text{BC}=\text{BC}$ (common sides) ......(2)
Since,$\text{AB}=\text{AC}$
$\frac{1}{2}\text{AB}=\frac{1}{2}\text{AC}$
As F and E are mid points of sides AB and AC respectively, so $\text{BF}=\text{CF}\ .....(3)$ From equation (1), (2), and (3)$\triangle\text{BFC}\cong\triangle\text{BEC}$
Hence $\text{FC}=\text{BE}$ Proved. View full question & answer→Question 84 Marks
BD and CE are bisectors of $\angle\text{B}$ and $\angle\text{C}$ of an isosceles $\triangle\text{ABC}$ with $\text{AB}=\text{AC}.$ Prove that $\text{BD}=\text{CE}.$
Answer
In $\triangle\text{ABC}$$\because\text{AB}=\text{AC}$
$\therefore\angle\text{ABC}=\angle\text{ACB}$ [Angle opposite to equal sides are equal]
$\Rightarrow\frac{1}{2}\angle\text{ABC}=\frac{1}{2}\angle\text{ACB}$
$\Rightarrow\angle\text{DBC}=\angle\text{ECB}$ $[$BD and CE bisects $\angle\text{B}$ and $\angle\text{C}]$
Now In $\triangle\text{DBC}$ and $\triangle\text{ECB}$$\angle\text{DBC}=\angle\text{ECB}$ [provedearlier]
$\angle\text{B}=\angle\text{C}$ [given]
$\text{BC}=\text{BC}$ [common]
By ASA congurence criterion $\triangle\text{DBC}\cong\triangle\text{ECB}$$\therefore\text{BD}=\text{CE}$ [c.p.c.t] View full question & answer→Question 94 Marks
O is any point in the interior of $\triangle\text{ABC}.$ Prove that
- $\text{AB}+\text{AC}>\text{OB}+\text{OC}$
- $\text{AB}+\text{BC}+\text{CA}>\text{OA}+\text{OB}+\text{OC}$
- $\text{OA}+\text{OB}=\text{OC}>\frac{1}{2}(\text{AB}+\text{BC}+\text{CA)}$
AnswerIt is given that, O is any point in the interior of $\triangle\text{ABC}$
We have to prove that.
- AB + AC > OB + OC produced BO to meet AC at D.
In $\triangle\text{ABC}$ we have
AB + AD > BD
⇒ AB + AD > OB + OD ......(1)
And in $\triangle\text{ODC}$ we have
OD + CD > OC ......(2)
Adding (1) & (2) we get.
AB + AD + OD + DC > OB + OD + OC
Hence AB + AC > OB + OC proved.
- We have to prove that AB + BC + CA > OA + OB + OC
From the first result we have
BC + BA > OA + OC .......(3)
And
CA + CB > OA + OB .......(4)
Adding above (4) equation.
2(AB + BC + AC) > 2(OA + OB + OC)
Hence AB + BC + CA > OA + OB + OC proved.
- We have to prove that $\text{OA}+\text{OB}+\text{OC}>\frac{1}{2}(\text{AB}+\text{BC}+\text{CA})$
In trianglesOAB, OBC and OCA we have
OA + OB > AB
OB + OC > BC
OC + OA > AC
Adding these three results.
2(OA + OB + OC) > AB + BC + AC
Hence $\text{OA}+\text{OB}+\text{OC}>\frac{1}{2}(\text{AB}+\text{BC}+\text{CA})$ Proved. View full question & answer→Question 104 Marks
AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See Fig). Show that the line PQ is perpendicular bisector of AB.
AnswerConsider the figure. 
We have AB is a line segment and P, Q are points on opposite sides of AB such that$\text{AP}=\text{BP}\ ....(\text{i)}$
$\text{AQ}=\text{BQ}\ ....(\text{ii)}$
We have to prove that PQ is perpendicular bisector of AB. Now consider $\triangle\text{PAQ}$ and $\triangle\text{PBQ},$ We have$\text{AP}=\text{BP}$ [From (i)]
$\text{AQ}=\text{BQ}$ [From (ii)]
And $\text{PQ}-\text{PQ}$ [Common site]$\triangle\text{PAQ}\simeq\triangle\text{PBQ}\ ....(\text{i})$(ii) [From SAS congruence]
Now, we can observe that APB and ABQ are isosceles triangles. [From (i) and (ii)]$\angle\text{PAB}=\angle\text{ABQ}$ and $\angle\text{QAB}=\angle\text{QBA}$
Now consider $\triangle\text{PAC}$ and $\triangle\text{PBC}$ C is the point of intersection of AB and PQ$\text{PA}=\text{PB}$ [From (i)]
$\angle\text{APC}=\angle\text{BPC}$ [From (ii)]
$\text{PC}=\text{PC}$ [common side]
So, from SAS congruency of triangle $\triangle\text{PAC}\cong\triangle\text{PBC}$$\text{AC}=\text{CB}$ and $\angle\text{PCA}=\angle\text{PBC}\ .....(\text{iv)}$[Corresponding parts of congruent triangles are equal]
And also, ACB is line segment$\angle\text{ACP}+\angle\text{BCP}=180^\circ$
$\angle\text{ACP}=\angle\text{PCB}$
$\angle\text{ACP}=\angle\text{PCB}=90^\circ<$
We have $\text{AC}=\text{CB}\Rightarrow\text{C}$ is the midpoint of AB From (iv) and (v) We can conclude that PC is the perpendicular bisector of AB Since C is a point on the line PQ, we can say that PQ is the perpendicular bisector of AB. View full question & answer→Question 114 Marks
In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
AnswerGiven that, in two right triangles one side and acute angle of one are equal to the corresponding side and angles of the other. 
We have to prove that the triangles are congruent. Let us consider two right triangles such that$\angle\text{B}=\angle\text{C}=90^\circ\ ....(\text{i})$
$\text{AB}=\text{DE}\ ....(\text{ii})$
$\angle\text{C}=\angle\text{F}$ From (iii)
Now observe the two triangles ABC and DEF$\angle\text{C}=\angle\text{F}$ [From (iii)]
$\angle\text{B}=\angle\text{E}$ [From (i)]
and $\text{AB}=\text{DE}$ [From (ii)] So, by AAS congruence criterion, we have $\triangle\text{ABC}\cong\triangle\text{DEF}$ Therefore, the two triangles are congruent Hence proved. View full question & answer→Question 124 Marks
In $\triangle\text{ABC},\angle\text{B}=35^\circ,\angle\text{C}=65^\circ$ and the bisector of $\angle\text{BAC}$ meets BC in P. Arrange AP, BP and CP in descending order.
AnswerIn angle
A + B + C = 180
A + 35 + 65 = 180
A = 180 - 100
A = 80
So $\angle\text{BAP}$ and $\angle\text{CAP}=\frac{80}{2}=40$
we know that side opposite to the greater angle is longer
In $\triangle\text{BAP}$
we know that side opposite to the greater angle is longer
so BP > AP
In $\triangle\text{CAP}$
ACP (65) > CAP (40)
so AP > CP
so BP > AP > CP
View full question & answer→Question 134 Marks
ABC is a triangle and D is the mid-point of BC. The perpendiculars from D to AB and AC are equal. prove that the triangle is isosceles.
Answer
In $\triangle\text{BDE}$ and $\triangle\text{CDF}$$\angle\text{BED}=\angle\text{DFC}=90^\circ$ [given]
$\text{DE}=\text{DF}$ [given]
$\text{BD}=\text{DC}$ [D is themidpoint]
By RHS congurence criterion $\triangle\text{BDE}\cong\triangle\text{CDF}$$\Rightarrow\angle\text{B}=\angle\text{C}$ [c.p.c.t]
$\Rightarrow\text{AB}=\text{AC}$ [Sides opposite to angles are equal]
Hence $\triangle\text{ABC}$ is isosceles. View full question & answer→Question 144 Marks
In a $\triangle\text{ABC},$ it is given that $\text{AB}=\text{AC}$ and the bisectors of B and C intersect at O. If M is a point on BO produced, prove that $\angle\text{MOC}=\angle\text{ABC}.$
AnswerGiven that in $\triangle\text{ABC},$$\text{AB}=\text{AC}$ and the bisector of $\angle\text{B}$ and $\angle\text{C}$ intersect at O. If M is a point on BO produced
We have to prove $\angle\text{MOC}=\angle\text{ABC}$ Since,$\text{AB}=\text{AC}$
ABC is isosceles$\angle\text{B}=\angle\text{C}\text{ (or})$
$\angle\text{ABC}=\angle\text{ACB}$
Now, BO and CO are bisectors of $\angle\text{ABC}$ and $\angle\text{ACB}$ respectively$\Rightarrow\text{ABO}=\angle\text{OBC}=\angle\text{ACO}=\angle\text{OCB}\\=\frac{1}{2}\angle\text{ABC}=\frac{1}{2}\angle\text{ACB}\ ...(\text{i)}$
We have, $\triangle\text{OBC}$$\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=180^\circ\ ....(\text{ii)}$
And also$\angle\text{BOC}+\angle\text{COM}=180^\circ\ ....(\text{iii)}$ [Straight angle]
Equating (ii) and (iii)$\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=\angle\text{BOC}+\angle\text{MOC}$
$\angle\text{OBC}+\angle\text{OCB}=\angle\text{MOC}$ [From (i)]
$2\Big(\frac{1}{2}\angle\text{ABC}\Big)=\angle\text{MOC}$ [From (i)]
$\angle\text{ABC}=\angle\text{MOC}$
Therefore, $\angle\text{MOC}-=\angle\text{ABC}$ View full question & answer→Question 154 Marks
In $\triangle\text{ABC},$ if $\angle\text{A}=40^\circ$ and $\angle\text{B}=60^\circ.$ Determine the longest and shortest sides of the triangle.
AnswerGiven that in $\triangle\text{ABC},\angle\text{A}=40^\circ$ and $\angle\text{B}=60^\circ$ 
We have to find longest and shortest side We know that, Sum of angles in a triangle 180°$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$40^\circ+60^\circ-\big((10)0^\circ\big)=180^\circ$
$\angle\text{C}=180^\circ$
Now,$\Rightarrow40^\circ<60^\circ<80^\circ=\angle\text{A}<\angle\text{B}<\angle\text{C}$
$\Rightarrow\angle\text{C}$ is greater angle and $\angle\text{A}$ is smaller angle.
Now, $\angle\text{A}<\angle\text{B}<\angle\text{C}$$\Rightarrow\text{BC}<\text{AAC}<\text{AB}$ [Side opposite to greater angle is larger and side opposite to smaller angle is smaller]
AB is longest and BC is smallest or shortest side. View full question & answer→Question 164 Marks
Find the measure of each exterior angle of an equilateral triangle.
AnswerGiven to find the measure of each exterior angle of an equilateral triangle consider an equilateral triangle ABC. We know that for an equilateral triangle$\text{AB}=\text{BC}=\text{CA}$ and $\angle\text{ABC}=\angle\text{BCA}=\text{CAB}=\frac{180^\circ}3=60^\circ\ ...(\text{i)}$
Now, Extend side BC to D, CA to E and AB to F. Here BCD is a straight line segment BCD = Straight angle = 180°$\angle\text{BCA}+\angle\text{ACD}=180^\circ$ [From (i)]
$60^\circ+\angle\text{ACD}=180^\circ $
$\angle\text{ACD}=120^\circ$
Similarly, we can find $\angle\text{FAB}$ and $\angle\text{FBC}$ also as 120° because ABC is an equilateral triangle$\angle\text{ACD}=\angle\text{EAB}-\angle\text{FBC}=120^\circ$
Hence, the median of each exterior angle of an equilateral triangle is 120°
View full question & answer→Question 174 Marks
Prove that each angle of an equilateral triangle is 60°.
AnswerGiven to prove each angle of an equilateral triangle is 60°. Let us consider an equilateral triangle ABC. Such that AB = BC = CA Now, AB = BC$\angle\text{A}=\angle\text{C}\ ....(\text{i})$ [Opposite angles to equal sides are equal]
And $\text{BC}=\text{AC}$$\angle\text{B}=\angle\text{A}\ ...(\text{ii)}$
From (i) and (ii), we get$\angle\text{A}=\angle\text{B}=\angle\text{C}\ ...(\text{iii})$
We know that Sum of angles in a triangle = 180$\angle\text{A}+\angle\text{B}+\angle\text{C}=180$
$\angle\text{A}+\angle\text{A}+\angle\text{A}=180$
$3\angle\text{A}=60$
$\angle\text{A}=60$
$\angle\text{A}=\angle\text{B}=\angle\text{C}=60$
Hence, each angle of an equilateral triangle is 60°. View full question & answer→Question 184 Marks
In the given figure, if $\text{AB}=\text{AC}$ and $\angle\text{B}=\angle\text{C}.$ Prove that $\text{PQ}=\text{CP}.$
AnswerIt is given that$\text{AB}=\text{AC},$ and $\angle\text{B}=\angle\text{C}$
We have to prove that $\text{BQ}=\text{CP}$ We basically will prove $\triangle\text{ABQ}\cong\triangle\text{ACP}$ to show $\text{BQ}=\text{CP}$ In $\triangle\text{ABQ}$ and $\triangle\text{ACP}$$\angle\text{B}=\angle\text{C}$ (Given)
$\text{AB}=\text{AC}$ (Given)
And $\angle\text{A}$ is common in both the triangles So all the properties of congruence are satisfied So $\triangle\text{ABQ}\cong\triangle\text{ACP}$ Hence $\text{BQ}=\text{CP}$ Proved. View full question & answer→Question 194 Marks
Is it possible to draw a triangle with sides of length 2cm, 3cm and 7cm?
AnswerGiven lengths of sides are 2cm, 3cm and 7cm To check whether it is possible to draw a triangle with the given lengths of sides We know that, A triangle can be drawn only when the sum of any two sides is greater than the third side. So, let's check the rule.$2+3\not>7\text{ or }2+3<7$
$2+7>3$
and $3+7>2$ Here $2+3\not=7$ So, the triangle does not exit.
View full question & answer→Question 204 Marks
ABC is a right angled triangle in which $\angle\text{A}=90^\circ$ and $\text{AB}=\text{AC}.$ and $\angle\text{B}$ and $\angle\text{C}.$
AnswerGiven that ABC is a right angled triangle such that $\angle\text{A}=90^\circ$ and $\text{AB}=\text{AC}$ since, $\text{AB}=\text{AC}$$\triangle\text{ABC}$ is also isosceles.
Therefore, we can say that $\triangle\text{ABC}$ is right angled isosceles triangle.$\angle\text{C}=\angle\text{B}$ and $\angle\text{A}=90^\circ \ ...(\text{i)}$
Now, we have sum of angled in a triangle = 180°$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$90^\circ+\angle\text{B}+\angle\text{B}=180^\circ$ [From (i)]
$2\angle\text{B}=180^\circ-90^\circ$
$\angle\text{B}=45^\circ$
Therefore, $\angle\text{B}=\angle\text{C}=45^\circ$
View full question & answer→Question 214 Marks
Two lines AB and CD intersect at O such that BC is equal and parellel to AD. prove that the lines AB and CD bisect at O.
AnswerIn $\triangle\text{AOD}$ and $\triangle\text{BOC}$$\angle\text{BCO}=\angle\text{ADO}$ [alternate angles]
$\angle\text{DAO}=\angle\text{CBO}$ [alternate angles]
$\text{BC}=\text{AD}$ [given]
By ASA conguence criterion $\triangle\text{AOD}\cong\triangle\text{BOC}$$\therefore\text{BO}=\text{OA}$ [c.p.c.t]
$\text{OC}=\text{OD}$ [c.p.c.t]
Therefore, AB and CD bisect at O.
View full question & answer→Question 224 Marks
In Fig. AB = AC and DB = DC, find the ratio $\angle\text{ABD}:\angle\text{ACD}.$
AnswerConsider the figure. 
Given, AB = AC, DB = DC and given to find the ratio$\angle\text{ABC}=\angle\text{ACD}$
Now, $\triangle\text{ABC}$ and $\triangle\text{DBC}$ are isosceles triangles since $\text{AB}=\text{AC}$ and $\text{DB}=\text{DC}$ respectively$\angle\text{ABC}=\angle\text{ACB}$ and $\angle\text{DBC}=\angle\text{DCB}$ [Angles opposite to equal sides are equal]
Now consider,$\angle\text{ABC}:\angle\text{ACD}$
$(\angle\text{ABC}-\angle\text{DBC}):(\angle\text{ACB}-\angle\text{DCB})$
$(\angle\text{ABC}-\angle\text{DBC}):(\angle\text{ABC}-\angle\text{DBC})\\ [\angle\text{ABC}=\angle\text{ABC}=\angle\text{ACB}\text{ and }\angle\text{DBC}=\angle\text{DBC]}$
$1:1$
$\text{ABC}:\text{ACD}=1:1$ View full question & answer→Question 234 Marks
If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.
AnswerGiven that the bisector of the exterior vertical angle of a triangle is parallel to the base and we have to prove that the triangle is isosceles. Let ABC be a triangle such that AD is the angular bisector of exterior vertical angle EAC and AD || BC. 
Let $\angle\text{EAD}= (\text{i}),$ $\angle\text{DAC}=(\text{ii}),$ $\angle\text{ABC}=(\text{iii})$ and $\angle\text{ACB}=(\text{iv})$ We have, (i) = (ii) $\big[$AD is a bisector of $\angle\text{EAC}\big]$ (i) = (iii) [Corresponding angles] and (ii) = (iv) [alternative angle] (iii) = (iv)$\text{AB}=\text{AC}$
Since, in $\triangle\text{ABC},$ two sides AB and AC are equal we can say that $\triangle\text{ABC}$ is isosceles triangle. View full question & answer→Question 244 Marks
In fig. PQRS is a square and SRT is an equilateral triangle. Prove that:
- PT = QT.
- $\angle\text{TQR}=15^\circ.$
AnswerGiven that PQRS is a square and SRT is an equilateral triangle. And given to prove that
- $\text{PT}=\text{QT}$
- $\angle\text{TQR}=15^\circ$
Now, PQRS is a square
$\text{PQ}=\text{QR}=\text{RS}=\text{SP}\ ....(\text{i)}$
and $\angle\text{SPQ}=\angle\text{PQR}=\angle\text{QRS}=\angle\text{RSP}=90^\circ=\text{tight angle}$
And also, SRT is an equilateral triangle.
$\text{SR}=\text{RT}=\text{TS}\ .....(\text{ii)}$
and $\angle\text{TSR}=\angle\text{SRT}=\angle\text{RTS}=60^\circ$
From (i) and (ii)
$\text{PQ}=\text{QR}=\text{SP}=\text{SR}=\text{RT}=\text{TS}\ ....(\text{iii)}$
And also,
$\angle\text{TSP}=\angle\text{TSR}+\angle\text{RSP}=60^\circ+90^\circ+150^\circ$
$\Rightarrow\angle\text{TSR}=\angle\text{TRQ}=150^\circ\ .....(\text{iv)}$
$\text{SP}=\text{RQ}$ [From (iii)]
So, by SAS congruence criterion we have
$\triangle\text{TSP}=\triangle\text{TRQ}$
$\text{PT}=\text{QT}$ [Corresponding parts of congruent triangles are equal]
Consider $\triangle\text{TQR}.$
$\text{QR}=\text{TR}$ [From (iii)]
$\triangle\text{TQR}$ is a isosceles triangle.
$\angle\text{QTR}=\angle\text{TQR}$ [angles opposite to equal sides]
Now,
Sum of angles in a triangle is equal to 180º.
$\Rightarrow\angle\text{QTR}+\angle\text{TQR}+\angle\text{TRQ}=180^\circ$
$\Rightarrow2\angle\text{TQR}+150^\circ=180^\circ$ [From (iv)]
$\Rightarrow2\angle\text{TQR}=180^\circ-150^\circ$
$\Rightarrow2\angle\text{TQR}=30^\circ$ $[\angle\text{TQR}=15^\circ]$
Hence proved. View full question & answer→Question 254 Marks
Prove that the sum of three altitudes of a triangle is less than the sum of its sides.
AnswerWe have to prove that the sum of three altitude of the triangle is less than the sum of its sides. In $\triangle\text{ABC}$ we have$\text{AD}\perp\text{BC},\text{BE}\perp\text{AC}$ and $\text{CF}\perp\text{AB}$
We have to prove $\text{AD}+\text{BE}+\text{CF}<\text{AB}+\text{BC}+\text{AC}$
As we know perpendicular line segment is shortest in length Since $\text{AD}\perp\text{BC}$ So $\text{AB}>\text{AD}\ .....(1)$ And $\text{AC}>\text{AD}\ ......(2)$
Adding (1) and (2) we get$\text{AB}+\text{AC}>\text{AD}+\text{AD}$
$\text{AB}+\text{AC}>2\text{AD}\ ......(3)$
Now $\text{BE}\perp\text{AC},$ so$\text{BC}+\text{BA}>\text{BE}+\text{BE}$
$\text{BC}+\text{BA}>2\text{BE}\ ......(4)$
And again $\text{CF}\perp\text{AB},$ this implies that$\text{AC}+\text{BC}>2\text{CF}\ .....(5)$
Adding (3) & (4) and (5) we have$(\text{AB}+\text{AC})+(\text{AB}+\text{BC})\\+(\text{AC}+\text{BC)}>2\text{AD}+2\text{BE}+2\text{CF}$
$\Rightarrow(\text{AB}+\text{BC}+\text{AC})>2(\text{AD}+\text{BE}+\text{CF})$
Hence $\text{AD}+\text{BE}+\text{CF}<\text{AB}+\text{BC}+\text{AC}$ Proved. View full question & answer→Question 264 Marks
CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that $\triangle\text{ADE}\cong\triangle\text{BCE}.$
AnswerWe have to prove that $\triangle\text{ADE}\cong\triangle\text{BCE}$ 
Given ABCD is a square So $\text{AB}=\text{BC}=\text{CD}=\text{AD}$ Now in $\triangle\text{EDC}$ is equilateral triangle. So $\text{DE}=\text{EC}=\text{CB}$ In $\triangle\text{AED}$ and $\triangle\text{CEB}$$\text{AD}=\text{BC}$ (Side of triangle)
$\text{DE}=\text{CE}$ (Side of equilateral triangle)
$\angle\text{ADE}=\text{ADC}+\angle\text{CDE}$
$=90+60$
$=150$
And,$\angle\text{BCE} =\angle\text{BCD}+\angle\text{DCE}$
$90+60$
$=150$
So $\angle\text{ADE}=\angle\text{BCE}$ Hence from SAS congruence $\triangle\text{ADE}\cong\triangle\text{BCE}$ Proved. View full question & answer→Question 274 Marks
Angles A, B, C of a triangle ABC are equal to each other. Prove that $\triangle\text{ABC}$ is equilateral.
AnswerGiven that angles A, B, C of a triangle ABC equal to each other. 
We have to prove that $\triangle\text{ABC}$ is equilateral We have, $\angle\text{A}=\angle\text{B}=\angle\text{C}$ Now,$\angle\text{A}=\angle\text{B}$
BC = AC [opposite sides to equal angles are equal] And $\angle\text{B}=\angle\text{C}$$\text{AC}=\text{AB}$
From the above we get$\text{AB}=\text{BC}=\text{AC}$
$\triangle\text{ABC}$ is equilateral. View full question & answer→Question 284 Marks
In a $\triangle\text{ABC},$ if $\angle\text{A}=120^\circ$ and $\text{AB}=\text{AC}.$ Find $\angle\text{A}$ and $\angle\text{C}.$
AnswerConsider a $\triangle\text{ABC}.$ 
Given Mat $\angle\text{A}=120^\circ$ and $\text{AB}=\text{AC}$ and given to find $\angle\text{B}$ and $\angle\text{C}.$ We can observe that $\triangle\text{ABC}$ is an isosceles triangle since $\text{AB}=\text{AC}$$\angle\text{B}=\angle\text{C}$ (i)
[Angles opposite to equal sides are equal] We know that sum of angles in a triangle is equal to 180°$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{B}=180^\circ$
$\Rightarrow120^\circ+2\angle\text{B}=180^\circ$
$\Rightarrow2\angle\text{B}=180^\circ-120^\circ$
$\Rightarrow\angle\text{B}=\angle\text{C}=30^\circ$ View full question & answer→Question 294 Marks
Determine the measure of each of the equal angles of a right-angled isosceles triangle. OR
ABC is a right-angled triangle in which $\angle\text{A}=90^\circ$ and $\text{AB}=\text{AC}.$ Find $\angle\text{A}$ and $\angle\text{C}.$
AnswerABC is a right angled triangle Consider on a right - angled isosceles triangle ABC such that$\angle\text{A}=90^\circ$ and $\text{AB}=\text{AC}$ since,
$\text{AB}=\text{AC}\Rightarrow\angle\text{C}=\angle\text{B}\ ....(\text{i)}$
[Angles opposite to equal sides are equal] Now, Sum of angles in a triangle = 180°$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow90^\circ+\angle\text{B}+\angle\text{B}=180^\circ$
$\Rightarrow2\angle\text{B}=90^\circ$
$\Rightarrow\angle\text{B}=45^\circ$
$\angle\text{B}=45^\circ,\angle\text{C}=45^\circ$
Hence, the measure of each of the equal angles of a right-angled Isosceles triangle Is 45°
View full question & answer→Question 304 Marks
Prove that the medians of an equilateral triangle are equal.
AnswerGiven, To prove the medians of an equilateral triangle are equal. Median: The line Joining the vertex and midpoint of opposite side. Now, consider an equilateral triangle ABC. Let D, E, F are midpoints of BC, CA and AB. 
Then, AD, BE and CF are medians of ABC. Now, D is midpoint of $\text{BC}\Rightarrow\text{BD}=\text{DC}=\frac{\text{BC}}{2}$ Similarly, $\text{CE}=\text{EA}=\frac{\text{AC}}{2}$$\text{AF}=\text{FB}=\frac{\text{AB}}{2}$
Since $\triangle\text{ABC}$ is an equilateral triangle$\Rightarrow\text{AB}=\text{BC}=\text{CA}\ ...(\text{i)}$
$\Rightarrow\text{BD}=\text{DC}=\text{CE}=\text{EA}=\text{AF}=\text{FB}\\=\frac{\text{BC}}{2}=\frac{\text{AC}}{2}=\frac{\text{AB}}{2}\ ...(\text{ii})$
And also, $\angle\text{ABC}=\angle\text{BCA}=\angle\text{CAB}=60^\circ\ ....(\text{iii)}$ Now, consider $\triangle\text{ABC}$ and $\triangle\text{BCE}\text{ AB}=\text{BC}$ [From (i)]$\text{BD}=\text{CE}$ [From (ii)]
Now, in $\triangle\text{TSR}$ and $\triangle\text{TRQ}$$\text{TS}=\text{TR}$ [From (iii)]
$\angle\text{ABD}=\angle\text{BCE}$ [From (iii)] $\big[\angle\text{ABD}$ and $\angle\text{ABC}$ and $\angle\text{BCE}$ and $\angle\text{BCA}$ are same $\big]$
So, from SAS congruence criterion, we have$\triangle\text{ABD}=\triangle\text{BCE}$
$\text{AD}=\text{BE}\ ...(\text{iv})$
[Corresponding parts of congruent triangles are equal] Now, consider $\triangle\text{BCE}$ and $\triangle\text{CAF},\text{BC}=\text{CA}$ [From (i)]$\angle\text{BCE}=\angle\text{CAF}$ [From (ii)]
$\big[\angle\text{BCE}$ and $\angle\text{BCA}$ and $\angle\text{CAF}$ and $\angle\text{CAB}$ are same $\big]$
$\text{CE}=\text{AF}$ [From (ii)]
So, from SAS congruence criterion, we have$\triangle\text{BCE}=\triangle\text{CAF}$
[Corresponding parts of congruent triangles are equal] From (iv) and (v), we have AD = BE = CF Median AD = Median BE = Median CF The medians of an equilateral triangle are equal. Hence proved View full question & answer→Question 314 Marks
Prove that the perimeter of a triangle is greater than the sum of its altitudes.
AnswerGiven, $\triangle\text{ABC}$ in which $\text{AD}\perp\text{BC},\text{ DE}\perp\text{AC }$and $\text{CF}\perp\text{AB}$ To prove, AD + BE + CF < AB + BC + AC Figure: 
Proof: We know that of all the segments that can be drawn to a given line, from a point not lying on it, the perpendicular distance i.e, the perpendicular line segment is the shortest. Therefore$\text{AD}\perp\text{BC}$
AB > AD and AC > AD
AB + AC > 2AD .... (i)
$\text{BE}\perp\text{AC}$
BA > BE and BC > BE
BA + BC > 2BE ... (ii)
$\text{CF}\perp\text{AB}$
CA > CF and CB > CF
CA + CB > 2CF ... (iii)
Adding (i), (ii) and (iii), we get AB + AC + BA + BC + CA + CB > 2AD + 2BE + 2CF 2AB + 2BC + 2CA > 2(AD + BE + CF) AB + BC + CA > AD + BE + CF The perimeter of the triangle is greater than that the sum of its altitudes. Hence proved View full question & answer→Question 324 Marks
The vertical angle of an isosceles triangle is 100º. Find its base angles.
AnswerConsider an isosceles $\triangle\text{ABC}$ such that $\text{AB}=\text{AC}$ Given that vertical angle A is 100º To find the base angles Since $\triangle\text{ABC}$ is isosceles$\angle\text{B}=\angle\text{C}$ [Angles opposite to equal sides are equal]
And also, Sum of interior angles of a triangle = 180°$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$(10)0^\circ+\angle\text{B}\angle\text{B}=180^\circ$
$2\angle\text{B}=180^\circ-(10)0^\circ$
$\angle\text{B}=40^\circ$
$\angle\text{B}=\angle\text{C}=40^\circ$
View full question & answer→Question 334 Marks
In $\triangle\text{ABC},$ side AB is produced to D so that $\text{BD}=\text{BC}.$ if $\angle\text{B}=60^\circ$ and $\angle\text{A}=70^\circ.$ Prove that: (i) AD > CD (ii) AD > AC
AnswerGiven that, in $\triangle\text{ABC},$ side AB is produced to D so that $\text{BD}=\text{BC}.$$\angle\text{B}=60^\circ,$ and $\angle\text{A}=70^\circ$
To prove, AD > CD AD > AC First join C and D We know that, Sum of angles in a triangle =180°$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$70^\circ+60^\circ+\angle\text{C}=180^\circ$
$\angle\text{C}=180^\circ-(130^\circ)=50^\circ$
$\angle\text{C}=50^\circ$
$\angle\text{ACB}=50^\circ\ ...(\text{i)}$
And also in $\triangle\text{BDC}$$\angle\text{DBC}=180^\circ-\angle\text{ABC}$ [ABD is a straight angle]
$180^\circ-60^\circ=120^\circ$
and also $\text{BD}=\text{BC}$ [given]$\angle\text{BCD}=\angle\text{BDC}$ [Angles opposite to equal sides are equal]
Now,$\angle\text{DBC}+\angle\text{BCD}+\angle\text{BDC}=180^\circ$ [Sum of angles in a triangle =180°]
$\Rightarrow120^\circ+\angle\text{BCD}+\angle\text{BCD}=180^\circ$
$\Rightarrow120^\circ+2\angle\text{BCD}=180^\circ$
$\Rightarrow2\angle\text{BCD}=180^\circ-120^\circ=60^\circ$
$\Rightarrow\angle\text{BCD}=30^\circ$
$\Rightarrow\angle\text{BCD}=\angle\text{BDC}=30^\circ\ ....(\text{ii)}$
Now, consider $\triangle\text{ADC}.$$\angle\text{BAC}\Rightarrow\angle\text{DAC}=70^\circ$ [given]
$\angle\text{BDC}\Rightarrow\angle\text{ADC}=30^\circ$ [From (ii)]
$\angle\text{ACD}=\angle\text{ACB}+\angle\text{BCD}$
= 50° + 30° [From (i) and (ii)] = 80° Now, $\angle\text{ADC}<\angle\text{DAC}<\angle\text{ACD}$$\text{AC}<\text{DC}<\text{AD}$ [Side opposite to greater angle is longer and smaller angle is smaller]
$\text{AD}>\text{CD}$ and $\text{AD}>\text{AC}$
Hence proved Or, We have,$\angle\text{ACD}>\angle\text{DAC}$ and $\angle\text{ACD}>\angle\text{ADC}$
$\text{AD}>\text{DC}$ and $\text{AD}>\text{AC}$ [Side opposite to greater angle is longer and smaller angle is smaller] View full question & answer→Question 344 Marks
ABCD is a square, X and Y are points on sides AD and BC respectively such that $\text{AY}=\text{BX}.$ prove that $\text{BY}=\text{AX}$ and $\angle\text{BAY}=\angle\text{ABX}.$
View full question & answer→Question 354 Marks
If perpendiculars from any point within an angle on its arms are congruents, prove that it lies on thebisector of that angle.
Answer
Here $\text{PM}=\text{PN}$ and $\angle\text{PMA}=\angle\text{PNA}=90^\circ$ In $\triangle\text{APM}$ and $\triangle\text{APN}$$\text{AP}=\text{AP}$ [common]
$\text{PN}=\text{PM}$ [given]
$\angle\text{PMA}=\angle\text{PNA}=90^\circ$ [given]
By RHS congurence criterion $\triangle\text{APN}\cong\triangle\text{APN}$$\therefore\angle\text{MAP}=\angle\text{NAP}$ [c.p.c.t]
Hence, AP is the bisector of $\angle\text{BAC}.$ View full question & answer→Question 364 Marks
In a $\triangle\text{ABC},$ if $\angle\text{B}=\angle\text{C}=45^\circ,$ which is the longest side?
AnswerGiven that in $\triangle\text{ABC},$ 
$\angle\text{B}=\angle\text{C}=45^\circ$
We have to find longest side We know that. Sum of angles in a triangle = 180°$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{A}+45^\circ+45^\circ=180^\circ$
$\angle\text{A}=180^\circ-(45^\circ+45^\circ)=180^\circ-90^\circ=90^\circ$
$\angle\text{A}=90^\circ$ View full question & answer→Question 374 Marks
P is a point on the bisector of an $\angle\text{ABC}.$ If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.
AnswerGiven that P is a point on the bisector of an $\angle\text{ABC},$ and $\text{PQ}\parallel\text{AB}.$ We have to prove that $\triangle\text{BPQ}$ is isosceles. 
Since, BP is bisector of $\angle\text{ABC}=\angle\text{ABP}=\angle\text{PBC}\ ....(\text{i)}$ Now,$\text{PQ}\parallel\text{AB}$
$\angle\text{BPQ}=\angle\text{ABP}\ ....(\text{ii)}$ [alternative angles]
From (i) and (ii), we get$\angle\text{BPQ}=\angle\text{PBC}\text{ (or}) \ \angle\text{BPQ}=\angle\text{PBQ}$
Now, In $\text{BPQ},$$\angle\text{BPQ}=\angle\text{PBQ}$
$\triangle\text{BPQ}$ is an isosceles triangle.
Hence proved. View full question & answer→Question 384 Marks
In a $\triangle\text{PQR},$ if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.
AnswerGiven that, In $\triangle\text{PQR},$ PQ = QR and L, M, N are midpoints of the sides PQ, QP and RP respectively and given to prove that LN = MN 
Here we can observe that $\triangle\text{PQR}$ is an isosceles triangle PQ = QR and $\angle\text{QPR}=\angle\text{QRP}\ ...(\text{i})$ And also, L and M are midpoints of PQ and QR respectively$\text{PL}=\text{LQ}=\text{QM}=\text{MR}=\frac{\text{PQ}}{2}=\frac{\text{QR}}{2}$
And also, $\text{PQ}=\text{QR}$ Now, consider $\triangle\text{LPN}$ and $\triangle\text{MRN},\text{LP}=\text{MR}$ [From - (2)]$\angle\text{LPN}=\angle\text{MRN}$ [From - (1)]
$\angle\text{QPR}$ and $\angle\text{LPN}$ and $\angle\text{QRP}$ and $\angle\text{MRN}$ are same.
PN = NR [N is midpoint of PR] So, by SAS congruence criterion, we have $\triangle\text{LPN}=\triangle\text{MRN}$ LN = MN [Corresponding parts of congruent triangles are equal] View full question & answer→Question 394 Marks
In Fig. it is given that $\text{RT}=\text{TS},\angle1=2\angle2$ and $\angle4=2\angle3.$ prove that $\triangle\text{RBT}\cong\triangle\text{SAT}.$
AnswerHere, $\angle1=2\angle2$ and $\angle4=2\angle3$ $\big[\therefore$ Exterior angle = sum of opposite interior angles $\big]$$\angle1=\angle4$ [vertically opposite angle]
$\therefore2\angle2=2\angle3$
$\Rightarrow\angle2=\angle3$
Now $\text{RT}=\text{TS}$ [given]$\Rightarrow\angle\text{TRS}=\angle\text{TSR}$ [Angle opposite to equal sides are equal]
$\therefore\angle\text{TRS}-\angle2=\angle\text{TSR}-\angle3$
$\Rightarrow\angle\text{TRB}=\angle\text{TSA}$
Now in $\triangle\text{RBT}$ and $\triangle\text{SAT}$$\angle\text{T}=\angle\text{T}$ [common]
$\angle\text{TRB}=\angle\text{TSA}$ [proved earlier]
$\text{RT}=\text{TS}$ [given]
By ASA conguence criterion $\triangle\text{RBT}\cong\triangle\text{SAT}$
View full question & answer→Question 404 Marks
In triangles ABC and CDE, if AC = CE, BC = CD, $\angle\text{A}=60^\circ,\angle\text{C}=30^\circ$ and $\angle\text{D}=90^\circ.$ Are two triangles congruent?
AnswerFor the triangles ABC and ECD, we have the following information and corresponding figure:$\text{AC}=\text{CE}$
$\text{BC}=\text{CD}$
$\angle\text{}A=60^\circ$
$\angle\text{C}=30^\circ$
$\angle\text{D}=90^\circ$
In triangles ABC and ECD, we have$\text{AC}=\text{EC}$
$\text{BC}=\text{CD}$
and $\angle\text{BAC}=\angle\text{CED}$ The SSA criteria for two triangles to be congruent are being followed. So both the triangles are congruent. View full question & answer→Question 414 Marks
PQR is a triangle in which PQ = PR and is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.
AnswerGiven that PQR is a triangle such that PQ = PR ant S is any point on the side PQ and ST || QR. 
To Prove, PS = PT Since, PQ = PR PQR is an isosceles triangle.$\angle\text{Q}=\angle\text{R}\text{ (or) }\angle\text{PQR}=\angle\text{PRQ}$
Now, $\angle\text{PST}=\angle\text{PQR}$ and $\angle\text{PTS}=\angle\text{PRQ}$ [Corresponding angles as ST parallel to QR] Since, $\angle\text{PQR}=\angle\text{PTS}$$\angle\text{PST}=\angle\text{PTS}$
Now, In $\triangle\text{PST},\angle\text{PST}=\angle\text{PTS}$$\triangle\text{PST}$ is an isosceles triangle
Therefore, $\text{PS}=\text{PT}$ View full question & answer→Question 424 Marks
In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
AnswerLet $\triangle\text{ABC}$ be isosceles such that $\text{AB}=\text{AC}.$$\angle\text{B}=\angle\text{C}$
Given that vertex angle A is twice the sum of the base angles B and C. i..e., $\angle\text{A}=2(\angle\text{B}=\angle\text{C})$$\angle\text{A}=2(\angle\text{B}+\angle\text{B})$
$\angle\text{A}=2(2\angle\text{B})$
$\angle\text{A}=4(\angle\text{B})$
Now, We know that sum of angles in a triangle = 180°$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$4\angle\text{B}+\angle\text{B}+\angle\text{B}=180^\circ$
$6\angle\text{B}=180^\circ$
$\angle\text{B}=30^\circ$
Since, $\angle\text{B}=4\angle\text{B}$$\angle\text{B}=\angle\text{C}=30^\circ$
And $\angle\text{A}=4\angle\text{B}$$\angle\text{A}=4\times30^\circ=120^\circ$
Therefore, angles of the given triangle are 120°, 30° and 30°.$=428$ and $\text{LB}=\text{LC}$ View full question & answer→Question 434 Marks
In Fig. the sides BA and CA have been produced such that: BA = AD and CA = AE. Prove that segment DE || BC.
AnswerGiven that, the sides BA and CA have been produced such that BA = AD and CA = AE and given to prove DE || BC Consider triangle BAC and DAE, 
We have BA = AD and CA = AE [given in the data] And also $\angle\text{BAC}=\angle\text{DAE}$ [vertically opposite angles] So, by SAS congruence criterion, we have$\angle\text{BAC}\simeq\angle\text{DAE}$
$\text{BC}=\text{DE}$ and $\angle\text{DEA}=\angle\text{BCA},\angle\text{EDA}=\angle\text{CBA}$
[Corresponding parts of congruent triangles are equal] Now, DE and BC are two lines intersected by a transversal that $\angle\text{DEA}=\angle\text{BCA}\text{ i}.\text{e}..$ Alternate angles are equal Therefore, DE, BC || BC. View full question & answer→