Question:
In the given figure, if x = y and AB = CB, then prove that AE = CD.
Solution:
Consider the triangles AEB and CDB.
$\angle E B A=\angle D B C$ (Common angle) ...(i)
Further, we have:
$\angle \mathrm{BEA}=180-\mathrm{y}$
$\angle \mathrm{BDC}=180-\mathrm{x}$
Since $\mathrm{x}=\mathrm{y}$, we have $:$
$180-\mathrm{x}=180-\mathrm{y}$
$\Rightarrow \angle \mathrm{BEA}=\angle \mathrm{BDC} \quad \ldots$ (ii)
$\mathrm{AB}=\mathrm{CB} \quad$ (Given) ...(iii)
From (i), (ii) and (iii), we have:
$\triangle B D C \cong \triangle B E A$ (AAS criterion)
$\therefore \mathrm{AE}=\mathrm{CD}(\mathrm{CPCT})$
Hence, proved.