Question

In the given figure, if x = y and AB = CB, then prove that AE = CD.

Answer 1

Consider the triangles AEB and CDB.

$\angle EBA=\angle DBC$ (Common angle) ...(i)

Further, we have:

$$\begin{gathered} \angle \mathrm{BEA}=180-\mathrm{y} \\ \angle \mathrm{BDC}=180-\mathrm{x} \\ \mathrm{Since}\mathrm{x}=\mathrm{y},\mathrm{we}\mathrm{have}: \\ 180-\mathrm{x}=180-\mathrm{y} \\ \Rightarrow \angle \mathrm{BEA}=\angle \mathrm{BDC}...\left(\mathrm{ii}\right) \end{gathered}$$

AB = CB (Given) ...(iii)
From (i), (ii) and (iii), we have:
$△BDC\cong △BEA$ (AAS criterion)
∴ AE = CD (CPCT)
Hence, proved.