$ABC$ 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 $\angle BAC = 72^o$.


Given:

$ABC$ is a triangle in which $\angle B = 2\angle C, D$ is a point on $BC$ such that $AD$ bisects $\angle BAC$ and $AB = CD$.

To do:

We have to prove that $\angle BAC = 72^o$.

Solution:

Draw the angular bisector of $\angle A B C$ which meets $AC$ in $P$.

Join $PD$

Let $\angle A C B=y$ and $\angle B A D=\angle D A C=x$

$\angle B=\angle A B C=2 \angle C=2 y$

$\angle B A C=2 x$

Therefore, $A D$ is the bisector of $\angle B A C$

In $\triangle B P C$

$\angle C B P=y$

Therefore, $B P$ is the bisector of $\angle A B C$

$\angle P C B=y$

$\angle C B P=\angle P C B=y$

This implies,

$P C=B P$

In $\triangle \mathrm{ABP}$ and $\triangle DCP$, we have,

$\angle \mathrm{ABP}=\angle \mathrm{DCP}=\mathrm{y}$

$\mathrm{AB}=\mathrm{DC}$

$\mathrm{PC}=\mathrm{BP}$

Therefore, by SAS axiom,

$\triangle \mathrm{ABP} \cong \triangle \mathrm{DCP}$

This implies,

$\angle B A P=\angle C D P$

$A P=D P$            (CPCT)

$\angle B A P=\angle C D P=2 x$

In $\triangle A B D$,

$\angle \mathrm{ABD}+\angle B A D+\angle A D B=180^{\circ}$

$\angle A D B+\angle A D C=180^{\circ}$

$2 x+2 y+y=180^{\circ}$

$2 y+3 y=180^{\circ}$

$5 y=180^{\circ}$

$y=\frac{180^{\circ}}{5}$

$y=36^{\circ}$

Therefore, $x=y=36^{\circ}$

$\angle A=\angle B A C=2 x=2 \times 36^{\circ}=72^{\circ}$

Therefore,

$\angle B A C=72^{\circ}$

Hence proved.

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Updated on: 10-Oct-2022

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