Question 1: In the adjoining figure, $AB = AC$ and $DB = DC$. Prove that $\angle ABD = \angle ACD$.

Consider $\triangle ABC$

Given $AB = AC \Rightarrow \angle ABC = \angle ACB$ (angles opposite to equal sides of a triangle are equal)

Now consider $\triangle BCD$

Given $DB = DC \Rightarrow \angle DBC = \angle DCB$ (angles opposite to equal sides of a triangle are equal)

Therefore $\angle ABC - \angle DBC = \angle ACB - \angle DCB$

$\Rightarrow \angle ABD = \angle ACD$. Hence proved.

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Question 2: In the adjoining figure, $AB = AC$. $BE$ and $CF$ are respectively the bisectors of $\angle ABC$ and $\angle ACB$. Prove that $\triangle EBC \cong \triangle FCB$.

Consider $\triangle ABC$

Given $AB = AC \Rightarrow \angle ABC = \angle ACB$ (angles opposite to equal sides of a triangle are equal)

Therefore we can also say that  $\angle FBC = \angle ECB$

Since $\angle ABC = \angle ACB$

$\Rightarrow \frac{1}{2}$ $\angle ABC =$ $\frac{1}{2}$ $\angle ACB$

$\Rightarrow \angle EBC = \angle FCB$

Now consider $\triangle EBC$ and $\triangle FCB$

$\angle EBC = \angle FCB$

$BC$ is common

$\angle FBC = \angle ECB$

Therefore $\triangle EBC \cong \triangle FCB$ (by A.S.A theorem). Hence proved.

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Question 3: If  $\triangle ABC$ is an isosceles triangle with $AB=AC$. Prove that perpendiculars from the vertices $B$ and $C$ to their opposite sides are equal.

Consider $\triangle ABC$

Given $AB = AC \Rightarrow \angle ABC = \angle ACB$ (angles opposite to equal sides of a triangle are equal)

Now consider $\triangle EBC$ and $\triangle DCB$

$\angle ABC = \angle ACB$

$BC$ is common

$\angle BEC = \angle CDB$ (right angles)

Therefore $\triangle EBC \cong \triangle DCB$ (by A.S.A theorem). Hence proved.

$\Rightarrow BD = CE$

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Question 4: In the adjoining figure, it is given that $\angle BAD = \angle BCE$ and $AB = BC$. Prove that $\triangle ABD \cong \triangle CBE$

Given $\angle BAD = \angle BCE$

and $\angle EOA = \angle DOC$ (vertically opposite angles)

$\Rightarrow \angle AEO = \angle CDO$

$\Rightarrow \angle CEB = \angle ADB$

Now consider $\triangle ABD$ and $\triangle CBE$

$\angle BAD = \angle BCE$

$AB = AC$ (given)

$\angle CEB = \angle ADB$

$\triangle ABD \cong \triangle CBE$ (by A.S.A theorem). Hence proved.

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Question 5: In $\triangle ABC, AB = AC$, and the bisector of $\angle ABC$ and $\angle ACB$ intersect at $O$. Prove that $BO = CO$  and the $\overrightarrow{OA}$ is the bisector of $\angle BAC$

Consider $\triangle ABC$

Given $AB = AC \Rightarrow \angle ABC = \angle ACB$ (angles opposite to equal sides of a triangle are equal)

$\Rightarrow \frac{1}{2}$ $\angle ABC$ $=$ $\frac{1}{2}$ $\angle ACB$

$\Rightarrow \angle OBC = \angle OCB$

$\Rightarrow \angle OBA = \angle OCA$

$\Rightarrow OB = OC$ (since sides opposite to equal angles are equal)

Now consider $\triangle AOB$ and $\triangle AOC$

$AB = AC$ (given)

$\angle OBA = \angle OCA$

$OB = OC$

Therefore $\triangle AOB \cong \triangle AOC$ (by S.A.S theorem)

$\Rightarrow \angle BAO = \angle CAO$

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Question 6: In the adjoining figure, it is given $AB = AC$ and $BD = EC$. Prove that $\triangle ABE \cong \triangle CAD$

Consider $\triangle ABC$

Given $AB = AC \Rightarrow \angle ABC = \angle ACB$ (angles opposite to equal sides of a triangle are equal)

$\Rightarrow \angle ABE = \angle ACD$

We have $BD = EC$

$\Rightarrow BD + DE = EC + DE$

$\Rightarrow BE = CD$

Consider $\triangle ABE$ and $\triangle ACD$

$AB = AC$ (given)

$\angle ABC = \angle ACB$

$BE = CD$

Therefore $\triangle ABE \cong \triangle ACD$ (by S.A.S. theorem)

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Question 7: In the adjoining figure, line $l$ is the bisector of $\angle QAP$ and $B$ is any point on $l$. $BP$ and $BQ$ are perpendiculars from $B$ to the arms of $A$. Show that: (i) $\triangle APB \cong \triangle AQB$ (ii) $BP = BQ$ or $B$ is equidistant from the arms of $\angle A$.

Since $l$ is the bisector of $\angle QAP$

$\Rightarrow \angle QAB = \angle PAB$

$\angle APB = \angle AQB$ (perpendicular lines)

$AB$ is common.

Therefore $\triangle AQB \cong \triangle APB$ (by A.S.A theorem)

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Question 8: In the adjoining figure $AD$ is a median and $BL, CM$ are perpendiculars drawn from $B$ and $C$ respectively on $AD$ and $AD$ produced. Prove that $BL = CM$

Consider $\triangle BDL$ and $\triangle CDM$

$\angle LDB = \angle CDM$ (vertically opposite angles)

$\angle BLD = \angle CMD$ (right angles)

$BD = DC (AD$ is the median $)$

Therefore $\triangle BDL \cong \triangle CDM$ (A.A.S theorem)

$\Rightarrow BL = CM$

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Question 9: In the adjoining figure, $BM \perp AC$ and $DN \perp AC$. Also $BM = DN$. Prove that $AC$ bisects $BD$

Consider $\triangle BMR$ and $\triangle DNR$

$\angle BMR = \angle DNR$ (right angles)

$\angle BRM = \angle DRN$ (vertically opposite angles)

and $BM = DN$ (given)

Therefor $\triangle BMR \cong \triangle DNR$ (by A.A.S theorem)

$\Rightarrow BR = DR$. Hence proved.

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Question 10: In the adjoining figure, $ABC$ is an isosceles triangle with $AB = AC. BD$ and $CE$ are two medians of the triangle. Prove that $BD = CE$.

Consider $\triangle ABC$

Given $AB = AC \Rightarrow \angle ABC = \angle ACB$ (angles opposite to equal sides of a triangle are equal)

Also, $AB = AC \Rightarrow \frac{1}{2} AB = \frac{1}{2} AC$

$\Rightarrow BE = DC$

Consider $\triangle BEC$ and $\triangle CDB$

$BC$ is common

$\angle ABC = \angle ACB$

$BE = DC$

Therefore $\triangle BEC \cong \triangle CDB$ (by S.A.S theorem)

Therefore $BD = CE$. Hence proved.

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Question 11: In the adjoining figure, $PS = PR$, $\angle TPS = \angle QPR$. Prove that $PT = PQ$

Consider $\triangle PSR$

Given $PS = PR \Rightarrow \angle PSR = \angle PRS$ (angles opposite to equal sides of a triangle are equal)

$\Rightarrow \angle PST = \angle PRQ$

Consider $\triangle PST$ and $\triangle PRQ$

$\angle PST = \angle PRQ$

$PS = PR$

$\angle TPS = \angle QPR$ (given)

Therefore $\triangle PST \cong \triangle PRQ$

Therefore $PT = PQ$

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Question 12: In the adjoining figure, $\triangle ABC$ and $\triangle DBC$ are two triangles on the same base $BC$ such that $AB = BC$ and $DB = DC$. Prove that $\angle ABD = \angle ACD$

Consider $\triangle ABC$

Given $AB = AC \Rightarrow \angle ABC = \angle ACB$ … … … … … (i) (angles opposite to equal sides of a triangle are equal)

Similarly in  $\triangle DBC$

Given $DB = DC \Rightarrow \angle DBC = \angle DCB$ … … … … … (ii) (angles opposite to equal sides of a triangle are equal)

Subtracting (ii) from (i) we get

$\angle ABC - \angle DBC = \angle ACB - \angle DCB$

$\angle ABD = \angle ACD$

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Question 13:  In the adjoining figure,  $AB \parallel CD$. $O$ is the midpoint of $AD$. Show that (i) $\triangle AOB \cong \triangle DOC$ (ii) $O$ is also the midpoint of $BC$

(i) Consider $\triangle AOB$ and $\triangle DOC$

$OA = OD$ (given)

Since $AB \parallel CD$

$\Rightarrow \angle ABO = \angle DCO$

also $\angle AOB = \angle DOC$ (vertically opposite angles)

Therefore $\triangle AOB \cong \triangle DOC$ (by A.A.S theorem)

(ii) Since $\triangle AOB \cong \triangle DOC$

$\Rightarrow OC = OB$

$\Rightarrow O$ is the midpoint of $BC$

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Question 14: In a $\triangle ABC$, it is given that $AB =AC$ and the bisectors of $\angle B$ and $\angle C$ intersect at $O$. If $M$ is a point on $BO$ produced, prove that $\angle MOC = \angle ABC$

Consider $\triangle ABC$

Given $AB = AC \Rightarrow \angle ABC = \angle ACB$ (angles opposite to equal sides of a triangle are equal)

$\Rightarrow \frac{1}{2} \angle ABC = \frac{1}{2} \angle ACB$

$\Rightarrow \angle OBC = \angle OCB$

In $\triangle OBC$, we have

$\angle MOC = \angle OBC + \angle OCB$

$\Rightarrow \angle MOC = \angle OBC + \angle OBC$

$\Rightarrow \angle MOC = 2 \angle OBC = \angle ABC$. Hence proved.

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Question 15: $P$ is a point on the bisector of an $\angle ABC$. If the line through $P$ parallel to $AB$ meets $BC$ at $Q$, prove that the $\triangle BPQ$ is isosceles.

Since $AB \parallel PQ$

$\angle ABP = \angle BPQ$ (alternate angles)

Also since $BP bisects \angle ABQ$

$\angle ABP = \angle QBP$

Hence $\angle QBP = \angle BPQ$

$\Rightarrow BQ = QP$ (sides opposite equal angles in a triangle are equal).

Hence $\triangle BQP$ is an isosceles triangle.

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Question 16: $ABC$ is a triangle in which $\angle B = 2 \angle C$. $D$ is a point on $BC$ such that $AD$ bisects $\angle BAC$ and $AB = CD$. Prove that $\angle BAC = 72^o$

In $\triangle ABC, \angle ABC = 2 \angle ACB$

Let $\angle ABC = 2y$ , where $\angle ACB = y$

$AD$ is the bisector of $\angle BAC$. So $\angle BAD = \angle CAD = x$

Let $BP$ be the bisector of $\angle ABC$.

In $\triangle BPC$, we have $\angle CBP = \angle BCP = y \Rightarrow BP = PC$

In $\triangle ABP$ and $\triangle DCP$, we have

$\angle ABP = \angle DCP = y$

$AP = DC$ (given) and $BP = PC$

Therefore $\triangle ABP \cong \triangle DCP$ (by S.A.S theorem)

$\Rightarrow \angle BAP = \angle CDP$ and $AP = DP$

$\Rightarrow \angle CDP = 2x$ and $\angle ADP = x$

In $\triangle ABD$, we have

$\angle ADC = \angle ABD + \angle BAD$

$\Rightarrow x+2x = 2y + x \Rightarrow x = y$

In $\triangle ABC$, we have

$\angle BAC + \angle ABC + \angle ACB = 180^o$

$\Rightarrow 2x+2y + y = 180^o$

$\Rightarrow 5x = 180^o$ $\Rightarrow x = 36^o$

Hence $\angle BAC = 2x = 72^o$

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