2 Marks Questions

Question 12 Marks

ABC is an isosceles triangle with AB = AC. Draw AP $\perp$ BC to show that $\angle$B = $\angle$C.

Answer

Given: $A B C$ is an isosceles triangle with $A B=A C$.
To Prove: $\angle B =\angle C$
Construction: Draw $A P \perp B C$
Proof: In right triangle APB and right triangle $APC.$
$AB = AC . . . .$ [given]
$AP = AP . . . .$[Common]
$\therefore \triangle APB \cong \triangle APC \ldots[RHS \text { rule] }$
$\therefore \angle ABP=\angle ACP \ldots[\text { c.p.c.t. }] $
$\therefore \angle B=\angle C$

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Question 22 Marks

$B E$ and $C F$ are two equal altitudes of a triangle $A B C$. Using $RHS$ congruence rule, prove that a triangle $A B C$ is isosceles.

Answer

Given: $BE $and $CF $are two equal altitudes of a triangle $ABC .$
To Prove: $\triangle A B C$ is isosceles.
Proof : In right $\triangle B E C$ and right $\triangle C F B$
side $B E=$ side $C F$...[Given]
$B C=C B$...[Common]
$\triangle B E C=\triangle C F B \ldots[B y \text { RHS rule }]$
$\therefore \angle BCE=\angle CBF \ldots[\text { c.p.c.t.] }$
$\therefore A B=A C$...[Sides opposite to equal angles of a triangle are equal]
$\therefore \triangle ABC$ is isosceles.

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Question 32 Marks

ABC is a right-angled triangle in which $\angle A =90^{\circ}$ and $AB = AC$. Find $\angle B$ and $\angle C$.

Answer

$ABC$ is a right triangle in which,

$AB = AC$
$\Rightarrow \angle C =\angle B$ [ opposite angles of equal side are equal]$ .....(i)$
We know that, in $ABC , \angle A +\angle B +\angle C =180^{\circ}$ [ Angle sum property of triangle]
$\Rightarrow 90^{\circ}+\angle B+\angle B=180^{\circ}\left[\angle A=90^{\circ} \text { (given) and } \angle B=C \text { (from eq. (i) }\right]$
$\Rightarrow 2 \angle B=180^{\circ}-90^{\circ}$
$\Rightarrow 2 \angle B=90^{\circ}$
$\Rightarrow \angle B=45^{\circ}$
Also $\angle C =45^{\circ}[$ As $\angle B =\angle C ]$

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Question 42 Marks

$ABC$ and $DBC$ are two isosceles triangles on the same base $BC$. Show that $\angle ABD = \angle ACD.$

Answer

Given : $A B C$ and $D B C$ are two isosceles triangles on the same base $B C$.
To Prove : $\angle ABD =\angle ACD$.
Proof : $A s A B C$ is an isosceles triangle on the base $B C$
$\therefore \angle ABC=\angle ACB \ldots(1)$
As $A B C$ is an isosceles triangle on the base $B C$
$\therefore \angle DBC=\angle DCB \ldots(2)$
Adding the corresponding sides of $(1)$ and $(2)$
$\angle ABC + \angle DBC = \angle ACB + \angle DCB$
$\Rightarrow \angle ABD = \angle ACD$

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Question 52 Marks

$ABC$ is a triangle in which altitudes BE and $CF$ to sides $AC$ and $AB$ are equal. Show that $\triangle \mathrm { ABE } \cong \triangle \mathrm { ACF }, AB = AC$ i.e. $\triangle$ABC is an isosceles triangle.

Answer

Given : $ABC$ is a triangle in which altitude $BE$ and $CF$ to side $AC$ and $AB$ are equal.
To Prove : $\triangle \mathrm { ABE } \cong \triangle \mathrm { ACF }$
i. $A B=A C$ i.e. $\triangle A B C$ is an isosceles triangle.
Proof: $BE = CF$...... [Given]
$\angle B A E=\angle C A F$...... [Common]
$\angle A F B=\angle A F C$........ [Each $90 { }^{\circ}$ ]
$\therefore \triangle ABE \cong \triangle ACF$ $\qquad$ [By $AAS$ property]
ii. $\triangle A B E \cong \triangle A C F \ldots$...... [As proved]
$\therefore AB = AC \ldots [c.p.c.t.]$
$\therefore \triangle ABC$ is an isosceles triangle.

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Question 62 Marks

$\triangle A B C, A D$ is perpendicular bisector of $B C$. Show that $\triangle A B C$ is an isosceles triangle in which $A B=A C$.

Answer

Given: $A D \perp B C$.
To prove : $A B C$ is an isosceles triangle in which $A B=A C$.
Proof: In $\triangle A D B$ and $\triangle A D C$,
$D B=D C \ldots[A s A D \perp$ bisector of $B C]$
$\angle \mathrm{ADB}=\angle \mathrm{ADC} \ldots\left[\right.$ Each $\left.90^{\circ}\right]$
$A D=A D \ldots[$ Common $]$
$\therefore \triangle \mathrm{ADB} \cong \triangle \mathrm{ADC} \ldots[$ By $SAS$ property]
$\therefore A B=A C \ldots$ [c.p.c.t]
$\therefore \triangle A B C$ is an isosceles triangle in which $A B=A C$.

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Question 72 Marks

In figure, $A C=A E, A B=A D$ and $\angle B A D=\angle E A C$. Show that $B C=D E$.

Answer

Given : $AC = AE, AB = AD$ and $\angle BAD = \angle EAC.$
To prove ; $BC = DE$
Proof : In $DABC$ and $DADE$
$AC = AE, AB = AD$ and $\angle BAD = \angle EAC ...[$Given]
$\therefore \angle BAD + \angle DAC = \angle DAC + \angle EAC .$..[Adding $\angle DAC $to both sides]
$\therefore \angle BAC = \angle DAE ...(1) $
$ AC = AE ... $ [Given]
$\angle BAC = \angle DAE ...$ [From$ (1)] $
$ AB = AD ... $[Given]
$\therefore DABC \cong DADE ...$ [By $SAS$ property]
$\therefore BC = DE ...$ [c.p.c.t.]

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Question 82 Marks

$A B C D$ is a quadrilateral in which $A D=B C$ and $\angle D A B=\angle C B A$ : Prove that:
i. $\triangle A B D \cong \triangle B A C$
ii. $B D=A C$
iii. $\angle ABD =\angle BAC$

Answer

In quadrilateral $ACBD$ , we have $AD = BC$ and $\angle DAB =\angle CBA$
i. In $\triangle ABC$ and $\triangle BAC$,
$A D=B C$ (Given)
$\angle DAB =\angle CBA$ (Given)
$AB = AB$ (Common)
$\triangle A B D \cong \triangle B A C \ldots[B y$ $SAS$ Congruence]
ii. Since $\triangle A B D \cong \triangle B A C$
$\Rightarrow BD = AC$ [By $C.P.C.T.]$
iii.Since $\triangle ABD ≅ \triangle BAC$
$\Rightarrow \angle ABD = \angle BAC$ [By $C.P.C.T.]$

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Question 92 Marks

In quadrilateral $A B C D$ (See figure). $A C=A D$ and $A B$ bisects $\angle A$. Show that $\triangle A B C \cong \triangle A B D$. What can you say about $B C$ and $B D$ ?

Answer

Given: In quadrilateral $ABCD, AC = AD$ and $AB$ bisects $\angle A.$
To prove: $\angle ABC \cong \triangle ABD$
Proof: In $\triangle A B C$ and $\triangle A B D$,
$A C=A D$ [Given]
$\angle B A C=\angle B A D[\because AB$ bisects $\angle A]$
$AB = AB$ [Common]
$\therefore \triangle ABC \cong \triangle ABD$ [By $SAS$ congruency]
Thus $BC = BD$ [By $C.P.C.T.]$

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Question 102 Marks

P is a point equidistant from two lines l and m intersecting at a point A (see figure). Show that $AP$ bisects the angle between them.

Answer

Given: $P$ is a point equidistant from two lines $I$ and $m$ intersect at a point $A$.
To Prove: AP bisects the angle between them. i.e. $\angle \mathrm{PAB}=\angle \mathrm{PAC}$
Proof: Let $P B$ and $P C$ be perpendiculars from $P$ on lines $I$ and $m$ respectively. Since $P$ is equidistant from lines $I$ and $m$.
Therefore, $P B=P C$
In $\triangle P A B$ and $\triangle P A C$, we have
$\mathrm{PB}=\mathrm{PC} \text { [Given] }$
$\angle \mathrm{PBA}=\angle \mathrm{PCA}\left[\text { Each equal to } 90^{\circ}\right. \text { ] }$
$\text { and, } \mathrm{PA}=\mathrm{PA}[\mathrm{Common}]$
$\triangle \mathrm{PAB} \cong \triangle \mathrm{PAC}[\mathrm{By} \mathrm{~RHS} \text { congruence criterion] }$
$\Rightarrow \angle \mathrm{PAB}=\angle \mathrm{PAC}$
Hence Proved

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Question 112 Marks

$A B$ is a line segment. $P$ and $Q$ are points on opposite sides of $A B$ such that each of them is equidistant from the points $A$ and $B$ (See Figure). Show that the line $P Q$ is the perpendicular bisector of $A B$.

Answer

$\text { In } \triangle \mathrm{PAQ} \text { and } \triangle \mathrm{PBQ}$
$\mathrm{AP}=\mathrm{BP}(\text { Given })$
$\mathrm{AQ}=\mathrm{BQ} \text { (Given) }$
$\mathrm{PQ}=\mathrm{PQ} \text { (Common) }$
So, $\triangle P A Q \cong \triangle P B Q$ ($SSS$ rule)
Therefore, $\angle \mathrm{APQ}=\angle \mathrm{BPQ}(\mathrm{CPCT})$.
Now let us consider $\triangle P A C$ and $\triangle P B C$.
You have: $A P=B P$ (Given)
$\angle \mathrm{APC}=\angle \mathrm{BPC}(\angle \mathrm{APQ}=\angle \mathrm{BPQ}$ proved above $)$
$P C=P C($ Common)
So, $\triangle P A C \cong \triangle P B C$ ($SAS$ rule)
Therefore, $A C=B C(C P C T) .......(i)$
$\angle A C P=\angle B C P(C P C T)$
and $\angle A C P+\angle B C P=180^{\circ}$ (Linear pair)
So, $2 \angle \mathrm{ACP}=180^{\circ}$
Or, $\angle A C P=90^{\circ} .......(ii)$
From $(i)$ and $(ii),$ we can easily conclude that $PQ$ is the perpendicular bisector of $A B$.

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Question 122 Marks

$AB$ is a line segment and line $l$ is its perpendicular bisector. If a point $P$ lies on $l$, show that $P$ is equidistant from $A$ and $B$.

Answer

Let $C$ be the mid-point of $AB.$
Clearly, line $l$ passes through $C$ is perpendicular to $AB.$
In $\triangle PCA$ and $\triangle PCB$, we have

$A C=B C[\because C$ is the mid-point of $A B]$
$PC = PC$ [common side]
$\angle P C A=\angle P C B$ [Each equal to $90^{\circ}$ as $\left.I \perp AB \right]$
So, by $SAS$ congruence rule, we obtain
$\triangle P C A \cong \triangle P C B$
$\Rightarrow PA=PB$

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Question 132 Marks

In Fig., $OA = OB$ and $OD = OC$. Show that
$i. \triangle AOD \cong \triangle BOC$
$ii. AD \| BC$

Answer

$i.$ You may observe that in $\triangle A O D$ and $\triangle B O C$,
$OA = OB ($Given$)$
$OD = OC$
Also, since $\angle AOD$ and $\angle BOC$ form a pair of vertically opposite angles, we have $\angle A O D=\angle B O C$
So, $\triangle A O D \cong \triangle B O C ($by the $\text{SAS}$ congruence rule$)$
$ii.$ In congruent triangles $\text{AOD}$ and $\text{BOC}$ , the other corresponding parts are also equal.
So, $\angle O A D=\angle O B C$ and these form a pair of alternate angles for line segments $A D$ and $B C$.
Therefore, $A D\| B C$

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