Question
In the adjoining figure, $\triangle\text{ABC}$ is an isosceles triangle in which AB = AC. Also, D is a point such that BD = CD. Prove that AD bisects $\angle\text{A}$ and $\angle\text{D.}$

Answer

Given: In $\triangle\text{ABC,}$
AB = AC.
D is point such that BD = CD.
AD, BD and CD are joined.
To prove: Ad bisects $\angle\text{A}$ and $\angle\text{D}$
Proof: In $\triangle\text{ABD}$ and $\triangle\text{CAD,}$
AD = AD (Common)
AB = AC (given)
BD = CD (given)
$\triangle\text{ABD}\cong\triangle\text{CAD}$ (SSS condition)
$\angle\text{BAD}=\angle\text{CAD}$ (c.p.c.t.)
and $\angle\text{BDA}=\angle\text{CDA}$ (c.p.c.t.)
Hence, AD is the bisector of $\angle\text{A}$ and $\angle\text{D}.$
Hence proved.

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