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$ABC$ 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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