NIMCET 2018 Question Paper with solutions

Given, D is mid-point of ΔABC right angled at A.

In Δ ABC , $\mathit{B}{\mathit{C}}^2$ = $\mathit{A}{\mathit{B}}^2+\mathit{A}{\mathit{C}}^2$ ... (i)
and in Δ ARD $\mathit{B}{\mathit{D}}^2$ = $\mathit{A}{\mathit{B}}^2+\mathit{A}{\mathit{D}}^2$ ... (ii)
On subtracting Eq. (ii) from Eq. (i), we get
$\mathit{B}{\mathit{C}}^2-\mathit{B}{\mathit{D}}^2$ = $\mathit{A}{\mathit{B}}^2+\mathit{A}{\mathit{C}}^2-\mathit{A}{\mathit{B}}^2-\mathit{A}{\mathit{D}}^2$
= $\mathit{A}{\mathit{C}}^2-\mathit{A}{\mathit{D}}^2$
= ${\left(2\mathit{A}\mathit{D}\right)}^2-\mathit{A}{\mathit{D}}^2$ [Since AC = 2AD]
= ${\left(2\mathit{A}\mathit{D}\right)}^2-\mathit{A}{\mathit{D}}^2$
= $4\mathit{A}{\mathit{D}}^2-\mathit{A}{\mathit{D}}^2$ = 3 $\mathit{A}{\mathit{D}}^2$