SSC CGL Tier II 12 Jan 2017 Paper 1

(2)${\mathit{A}\mathit{C}}^2+{\mathit{C}\mathit{B}}^2={\mathit{A}\mathit{B}}^2$
$\Rightarrow 2{\mathit{B}\mathit{C}}^2={\left(\mathit{A}\mathit{D}+\mathit{D}\mathit{B}\right)}^2$
$\Rightarrow 2{\mathit{B}\mathit{C}}^2={\mathit{A}\mathit{D}}^2+{\mathit{D}\mathit{B}}^2+2\mathit{A}\mathit{D}.\mathit{B}\mathit{D}.....\left(\mathit{i}\right)$
ΔCEB and ΔCED are right angles.
CD${}^2$ = CE${}^2$ + ED${}^2$
and, BC${}^2$ = CE${}^2$ + BE${}^2$
BC${}^2$– CD${}^2$ = BE${}^2$ – DE${}^2$
= (BE + DE) (BE – DE)
= (AE + DE) (BE – DE)
= AD . BD ..... (ii)
∴From equations (i) and (ii)
AD${}^2$ + DB${}^2$ = 2CD${}^2$