Question

How to prove areas of similar triangles theorem

Answer 1

Draw $AM\perp BC\&PN\perp QR$

and $\Delta ABC\sim \Delta PQR$

$\begin{array}{cc}\frac{ar\left(\Delta ABC\right)}{ar\left(\Delta PQR\right)}=\frac{\frac{1}{2}\times BC\times AM}{\frac{1}{2}\times QR\times PN}& \\ Since,\Delta ABC\sim \Delta PQR& \\ \therefore \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}& \\ In\Delta ABM\&\Delta PQN& \\ \angle B=\angle Q& \left[\because \Delta ABC\sim \Delta PQR\right]\\ \angle M=\angle N={90}^{\circ }& \\ \therefore \Delta ABM\sim \Delta PQN& \\ \therefore \frac{AB}{PQ}=\frac{AM}{PN}& \\ \therefore \frac{ar\left(\Delta ABC\right)}{ar\left(\Delta PQR\right)}=\frac{AB}{PQ}\times \frac{AB}{PQ}={\left(\frac{AB}{PQ}\right)}^2={\left(\frac{BC}{QR}\right)}^2={\left(\frac{AC}{PR}\right)}^2& \end{array}$