SSC CHSL 15 Nov 2015 Shift 2 Paper

Given : In $∆\mathit{A}\mathit{B}\mathit{C},\mathit{A}\mathit{D}\perp \mathit{B}\mathit{C}$ and ${\mathit{A}\mathit{D}}^2=\mathit{B}\mathit{D}\times \mathit{D}\mathit{C}$
To find : $\angle \mathit{B}\mathit{A}\mathit{C}$
Solution : In right $∆\mathit{A}\mathit{D}\mathit{B}$ and $∆\mathit{A}\mathit{D}\mathit{C},$ if we apply Pythagoras Theorem,
⇒ ${\left(\mathit{A}\mathit{B}\right)}^2={\left(\mathit{A}\mathit{D}\right)}^2+{\left(\mathit{B}\mathit{D}\right)}^2$ ---(i)
and ⇒ ${\left(\mathit{A}\mathit{C}\right)}^2={\left(\mathit{A}\mathit{D}\right)}^2+{\left(\mathit{D}\mathit{C}\right)}^2$ ------(ii)
Adding equations (i) and (ii), we get :
⇒ ${\left(\mathit{A}\mathit{B}\right)}^2+{\left(\mathit{A}\mathit{C}\right)}^2=2{\left(\mathit{A}\mathit{D}\right)}^2$ $+{\left(\mathit{B}\mathit{D}\right)}^2+{\left(\mathit{D}\mathit{C}\right)}^2$
⇒ ${\left(\mathit{A}\mathit{B}\right)}^2+{\left(\mathit{A}\mathit{C}\right)}^2=2\left(\mathit{B}\mathit{D}\right)\left(\mathit{D}\mathit{C}\right)$ $+{\left(\mathit{B}\mathit{D}\right)}^2+{\left(\mathit{D}\mathit{C}\right)}^2$ [Given]
⇒ ${\left(\mathit{A}\mathit{B}\right)}^2+{\left(\mathit{A}\mathit{C}\right)}^2={\left(\mathit{B}\mathit{D}+\mathit{D}\mathit{C}\right)}^2$
⇒ ${\left(\mathit{A}\mathit{B}\right)}^2+{\left(\mathit{A}\mathit{C}\right)}^2={\left(\mathit{B}\mathit{C}\right)}^2$
Hence, $∆\mathit{A}\mathit{B}\mathit{C}$ is a right triangle right angled at A.
∴ ∠BAC = 90°