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.
$AB = AC.$
$D$ 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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