Question 13 Marks
In the adjoining figure, $AB > AC$ and D is any point on BC. Show that : $AB > AD$.
Answer
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Question 23 Marks
In the given figure, $AB > AC$. If BO and CO are the bisectors of $\angle B$ and $\angle C$ respectively, prove that $BO > CO$.
Answer
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Question 33 Marks
In the given figure, AD bisects $\angle A$. If $\angle B =60^{\circ}, \angle C =$ $40^{\circ}$, then arrange $AB , BD$ and DC in ascending order of their lengths.
Answer
View full question & answer→$BD = DC < AB$
Question 43 Marks
The sides AB and AC of $\triangle ABC$ are produced to D and E respectively and the bisectors of $\angle CBD$ and $\angle BCE$ meet at O . If $AB > AC$, prove that $OC > OB$.
Answer
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Question 53 Marks
In the adjoining figure, $\triangle ABC$ is equilateral and D is any point on AC.
Prove that :
(i) $BD > AD$
(ii) $BD > DC$.
Prove that :
(i) $BD > AD$
(ii) $BD > DC$.
Answer
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Question 63 Marks
In the adjoining figure, O is the centre of a circle, XY is a diameter and XZ is a chord. Prove that $XY > XZ$.
Answer
View full question & answer→[Hint. Join OZ. Then, $O X+O Z>X Z \Rightarrow X Y>X Z$
Question 73 Marks
In $\triangle ABC , D$ is any point on BC .
Prove that: $AB + BC + AC >2 AD$.
Prove that: $AB + BC + AC >2 AD$.
Answer
View full question & answer→[Hint. $A B+B D>A D$ and $A C+C D>A D$.]
Question 83 Marks
In $\triangle PQR , \angle P =50^{\circ}$ and $\angle R =70^{\circ}$.
Name : (i) the shortest side
(ii) the longest side of the triangle.
Name : (i) the shortest side
(ii) the longest side of the triangle.
Answer
View full question & answer→(i) QR
(ii) PQ
(ii) PQ
Question 93 Marks
In the given figure, side CA of $\triangle ABC$ has been produced to E . If $AC = AD = BD ; \angle ACD =46^{\circ}$ and $\angle BAE =x^{\circ}$; find the value of $x$.
Answer
View full question & answer→$x=69$
Question 103 Marks
If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.
Answer
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Question 113 Marks
In the given figure, $AB \| CD$ and $CA = CE$. Find the values of $x, y$ and $z$.
Answer
View full question & answer→$x=36, y=68, z=44$
Question 123 Marks
In the given figure, $AB = AC ; \angle A =50^{\circ}$ and $\angle ACD =15^{\circ}$. Show that $BC = CD$.
Answer
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Question 133 Marks
In each of the following figures, find the value of $x$ :
Answer
View full question & answer→(i) $x=40$
(ii) $x=34$
(ii) $x=34$
Question 143 Marks
In each of the following figures, find the value of $x$ :
Answer
View full question & answer→(i) $x=110$
(ii) $x=55$
(iii) $x=20$
(ii) $x=55$
(iii) $x=20$
Question 153 Marks
In the given figure, $\triangle ABC$ is an equilateral triangle and BC is produced to D such that $BC = CD$. Prove that $AD \perp AB$.
Answer
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Question 163 Marks
In the given figure, AD is the internal bisector of $\angle A$ and $CE \| DA$. If CE meets BA produced at E , prove that $\triangle CAE$ is isosceles.
Answer
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Question 173 Marks
In the given figure, $AB = AC$ and side BA has been produced to D . If AE is the bisector of $\angle CAD$, prove that $AE \| BC$.
Answer
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Question 183 Marks
Prove that the bisectors of the base angles of an isosceles triangle are equal.
Answer
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Question 193 Marks
In the given figure, $AB = AC$. If BO and CO , the bisectors of $\angle B$ and $\angle C$ respectively meet at O and BC is produced to D , prove that $\angle BOC =\angle ACD$.
Answer
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Question 203 Marks
In the given figure, $AB = AC ; D$ is the mid-point of BC ; $DP \perp BA$ and $DQ \perp CA$.
Prove that:
(i) $DP = DQ$
(ii) $AP = AQ$
(iii) AD bisects $\angle A$.
Prove that:
(i) $DP = DQ$
(ii) $AP = AQ$
(iii) AD bisects $\angle A$.
Answer
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Question 213 Marks
In a $\triangle ABC , BC = AC$ and $\angle B =64^{\circ}$, find $\angle C$.
Answer
View full question & answer→$\angle C =52$
Question 223 Marks
In the given figure, $AB = BC$ and $AC = CD$. Show that : $\angle BAD : \angle ADB =3: 1$.
Answer
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Question 233 Marks
In the given figure, $CA = CD = BD ; \angle DBC =35^{\circ}$ and $\angle DCA=x^{\circ}$. Find the value of $x$.
Answer
View full question & answer→$x=40$
Question 243 Marks
In a $\triangle A B C, A B=A C$ and $\angle A=50^{\circ}$, find $\angle B$ and $\angle C$.
Answer
View full question & answer→$\angle B =\angle C =65$
Question 253 Marks
In the given figure, the line segments AB and CD intersect at a point M in such a way that $AM = MD$ and $CM = MB$. Prove that, $AC = BD$ but AC may not be parallel to BD.
Answer
View full question & answer→[Hint. $\triangle M A C \cong \triangle M D B$. So, $\angle M D B=\angle M A C$.Thus, $\angle M D B \neq \angle M C A$.]
Question 263 Marks
In the given figure, ABCD is a square and $P , Q , R$ are points on $A B, B C$ and $C D$ respectively such that $AP = BQ = CR$ and $\angle PQR =90^{\circ}$.
Prove that :
(i) $PB = QC$
(ii) $PQ = QR$
(iii) $\angle QPR =45^{\circ}$
Prove that :
(i) $PB = QC$
(ii) $PQ = QR$
(iii) $\angle QPR =45^{\circ}$
Answer
View full question & answer→[Hint.$\begin{array}
{l}A B=B C \text { and } A P=B Q \Rightarrow A B-A P=B C-B Q \Rightarrow P B=Q C \\\text { Ext. } \angle R Q B=\angle Q C R+\angle Q R C=90^{\circ}+\angle Q R C \\\text { Also }, \angle R Q B=90^{\circ}+\angle P Q B \\\therefore \angle Q R C=\angle P Q B\end{array}$
Now, show that $\triangle P B Q \equiv \triangle Q C R$.]
{l}A B=B C \text { and } A P=B Q \Rightarrow A B-A P=B C-B Q \Rightarrow P B=Q C \\\text { Ext. } \angle R Q B=\angle Q C R+\angle Q R C=90^{\circ}+\angle Q R C \\\text { Also }, \angle R Q B=90^{\circ}+\angle P Q B \\\therefore \angle Q R C=\angle P Q B\end{array}$
Now, show that $\triangle P B Q \equiv \triangle Q C R$.]
Question 273 Marks
Equilateral triangles ABD and ACE are drawn on the sides $A B$ and $A C$ of $\triangle A B C$ as shown in the figure. Prove that :
(i) $\angle DAC =\angle EAB$
(ii) $DC = BE$.
(i) $\angle DAC =\angle EAB$
(ii) $DC = BE$.
Answer
View full question & answer→(Hint.
(i) $\angle D A C=60^{\circ}+\angle B A C$ and $\angle E A B=60^{\circ}+\angle B A C$.
(ii) Join DC and BE and show that $\triangle D A C \equiv \triangle B A E$.]
(i) $\angle D A C=60^{\circ}+\angle B A C$ and $\angle E A B=60^{\circ}+\angle B A C$.
(ii) Join DC and BE and show that $\triangle D A C \equiv \triangle B A E$.]
Question 283 Marks
In the given figure, ABCD is a square and $\triangle PAB$ is an equilateral triangle.
(i) Prove that $\triangle APD \cong \triangle BPC$.
(ii) Show that $\angle DPC =15^{\circ}$.
(i) Prove that $\triangle APD \cong \triangle BPC$.
(ii) Show that $\angle DPC =15^{\circ}$.
Answer
View full question & answer→[Hint.
(i) $\angle P A D=90^{\circ}+60^{\circ}=150^{\circ}$ and $\angle P B C=90^{\circ}+60^{\circ}=150^{\circ}$.
(ii) $\angle P A D=150^{\circ}$ and $A P=A D \Rightarrow \angle A P D=\angle A D P=15^{\circ}$.]
(i) $\angle P A D=90^{\circ}+60^{\circ}=150^{\circ}$ and $\angle P B C=90^{\circ}+60^{\circ}=150^{\circ}$.
(ii) $\angle P A D=150^{\circ}$ and $A P=A D \Rightarrow \angle A P D=\angle A D P=15^{\circ}$.]
Question 293 Marks
In the given figure : $AY \perp ZY$ and $BY \perp XY$ such that $AY = ZY$ and $BY = XY$.
Prove that $AB = ZX$.
Prove that $AB = ZX$.
Answer
View full question & answer→[Hint. $\triangle A Y B \cong \triangle Z Y X$.
Note that $\angle A Y B=90^{\circ}+\angle A Y X$ and $\angle Z Y X=90^{\circ}+\angle A Y X$. 1
Note that $\angle A Y B=90^{\circ}+\angle A Y X$ and $\angle Z Y X=90^{\circ}+\angle A Y X$. 1
Question 303 Marks
In the given figure : $\angle ABD =\angle EBC , BD = BC$ and $\angle ACB =\angle EDB$. Prove that $AB = BE$.
Answer
View full question & answer→[Hint. $\angle E B D=\angle A B E+\angle A B D$ and $\angle A B C=\angle A B E+\angle E B C$.
So, $\angle E B D=\angle A B C$.
Now, show that $\triangle A C B \cong \triangle E D B$.]
So, $\angle E B D=\angle A B C$.
Now, show that $\triangle A C B \cong \triangle E D B$.]
Question 313 Marks
Show that the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal.
Answer
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Question 323 Marks
In a $\triangle A B C, B C=A C$ and $\angle B=64^{\circ}$, find $\angle C$.
Answer
View full question & answer→$\angle C =52^{\circ}$
Question 333 Marks
In a $\triangle A B C, A B=A C$. If the bisectors of $\angle B$ and $\angle C$ meet $A C$ and $A B$ at points $D$ and $E$ respectively, show that :
(i) $\triangle DBC \cong \triangle ECB$
(ii) $BD = CE$.
(i) $\triangle DBC \cong \triangle ECB$
(ii) $BD = CE$.
Answer
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Question 343 Marks
In a $\triangle A B C, A B=A C$ and $\angle A=50^{\circ}$, find $\angle B$ and $\angle C$.
Answer
View full question & answer→$\angle B =\angle C =65^{\circ}$
Question 353 Marks
In an isosceles triangle, prove that the altitude from the vertex bisects the base.
Answer
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Question 363 Marks
If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.
Answer
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Question 373 Marks
If the altitude from one vertex of a triangle bisects the opposite side, prove that the triangle is isosceles.
Answer
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Question 383 Marks
In the given figure, find the value of $a+b+c+d+e+f$.
Answer
View full question & answer→$360^{\circ}$
Question 393 Marks
In $\triangle ABC , \angle B =35^{\circ}, \angle C =65^{\circ}$ and the bisector AD of $\angle BAC$ meets BC at D . Arrange the sides $AD , BD$ and CD in ascending order of their lengths.
Answer
View full question & answer→$BD > AD > CD$
Question 403 Marks
In the given figure, $\angle ABC =66^{\circ}, \angle DAC =38^{\circ}$. CE is perpendicular to $A B$ and $A D$ is perpendicular to $B C$. Prove that: $CP > AP$
Answer
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Question 413 Marks
In the given figure, $AD = AB$ and AE bisects $\angle A$. Prove that:
(i) $BE = ED$
(ii) $\angle ABD >\angle BCA$.
(i) $BE = ED$
(ii) $\angle ABD >\angle BCA$.
Answer
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Question 423 Marks
If $O$ is any point inside $\triangle A B C$, prove that $\angle B O C>\angle A$.
Answer
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Question 433 Marks
In the adjoining quadrilateral $ABCD , AB$ is the longest side and DC is the shortest side. Prove that:
(i) $\angle C >\angle A$
(ii) $\angle D >\angle B$
(i) $\angle C >\angle A$
(ii) $\angle D >\angle B$
Answer
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Question 443 Marks
In the adjoining figure, $O$ is the centre of a circle, $X Y$ is a diameter and XZ is a chord. Prove that $XY > XZ$.
Answer
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Question 453 Marks
In the adjoining figure, $AC > AB$ and AD is the bisector of $\angle A$. Show that : $\angle ADC >\angle ADB$.
Answer
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Question 463 Marks
In the adjoining figure, $AB > AC$ and D is any point on $B C$. Show that $: A B>A D$.
Answer
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Question 473 Marks
In the given figure, side $A B$ of $\triangle A B C$ is produced to $D$ such that $B D=B C$.
If $\angle A=60^{\circ}$ and $\angle B=50^{\circ}$, prove that:
(i) $AD > CD$
(ii) $AD > AC$
If $\angle A=60^{\circ}$ and $\angle B=50^{\circ}$, prove that:
(i) $AD > CD$
(ii) $AD > AC$
Answer
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Question 483 Marks
In the adjoining figure, $AB = AC$. If $DB \perp BC$ and EC $\perp BC$, prove that :
(i) $BD = CE$
(ii) $AD = AE$.
(i) $BD = CE$
(ii) $AD = AE$.
Answer
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Question 493 Marks
In the given figure, AD is the internal bisector of $\angle A$ and $CE \| DA$. If CE meets BA produced at E , prove that $\triangle CAE$ is isosceles.
Answer
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Question 503 Marks
In the given figure, $A B=A C$ and side $B A$ has been produced to D . If AE is the bisector of $\angle CAD$, prove that $AE \| BC$.
Answer
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