Question
In the given figure; $A B=B C$ and $A D=E C$. Prove that: $BD = BE$.
✓
Answer
$\text { In } \triangle \mathrm{ABC}$
$ \mathrm{AB}=\mathrm{BC} \ldots . . . ($given$)$
$ \Rightarrow \angle \mathrm{BCA}=\angle \mathrm{BAC} \ldots . . . ($Angles opposite to equal sides are equal$)$
$ \Rightarrow \angle B C D=\angle B A E \ldots . \text { (i) }$
Given, $\mathrm{AD}=\mathrm{EC}$
$ \Rightarrow \mathrm{AD}+\mathrm{DE}=\mathrm{EC}+\mathrm{DE} \ldots ($Adding $\mathrm{DE}$ on both sides$)$
$ \Rightarrow \mathrm{AE}=\mathrm{CD} \ldots . . . . . \text { (ii) }$
Now, in triangles $A B E$ and $C B D$,
$\mathrm{AB}=\mathrm{BC} \ldots . . . ($given$)$
$ \angle \mathrm{BAE}=\angle \mathrm{BCD} \ldots . .[$ From $(i)] $
$ \mathrm{AE}=\mathrm{CD} \ldots . . .[$ From $(ii)] $
$ \Rightarrow \triangle \mathrm{ABE} \cong \triangle \mathrm{CBD}$
$ \Rightarrow \mathrm{BE}=\mathrm{BD} \ldots . .(\mathrm{cpct})$
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