Question
In the following figure, $\angle A=\angle C$ and $A B=B C$. Prove that $\triangle A B D \cong \triangle C B E$.
✓
Answer
In triangles $A O E$ and $C O D$,
$\angle A=\angle C \dots...$(given$)$
$\angle \mathrm{AOE}=\angle \mathrm{COD} \dots...($vertically opposite angles$)$
$\begin{aligned} & \therefore \angle \mathrm{A}+\angle \mathrm{AOE}=\angle \mathrm{C}+\angle \mathrm{COD} \\ & \Rightarrow 180^{\circ}-\angle \mathrm{AEO}=180^{\circ}-\angle \mathrm{CDO}\end{aligned}$
$\Rightarrow \angle \mathrm{AEO}=\angle \mathrm{CDO} \dots....(i)$
Now,$\angle \mathrm{AEO}+\angle \mathrm{OEB}=180^{\circ} \quad \ldots...($linear pair$)$
And, $\angle C D O+\angle O D B=180^{\circ}....$(linear pair$)$
$\therefore \angle \mathrm{AEO}+\angle \mathrm{OEB}=\angle \mathrm{CDO}+\angle \mathrm{ODB}$
$\Rightarrow \angle \mathrm{OEB}=\angle \mathrm{ODB} \dots....[$ Using $(i) ]$
$\Rightarrow \angle \mathrm{CEB}=\angle \mathrm{ADB}$ ....(ii)
Now, in $\triangle \mathrm{ABD}$ and $\triangle \mathrm{CBE}$,
$\angle A=\angle C \dots.... ($given$)$
$\angle \mathrm{ADB}=\angle \mathrm{CEB}$ ...[ From (ii) ]
$A B=B C \dots....$(given$)$
$\Rightarrow \triangle \mathrm{ABD} \cong \triangle \mathrm{CBE} \dots...($by $\text{AAS}$ congruence criterion$).$
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