Question

Prove that the areas of two similar triangles are in the ratio of the squares of their corresponding: [4 MARKS]
(i) altitudes
(ii) Medians
(iii) Angle bisector segments

Answer 1

Application of theorems: 2 Marks
Calculation: 2 Marks
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$\begin{array}{ccc}\left(i\right)Proof:& \frac{ar\left(\Delta ABC\right)}{ar(\Delta PQR}=\frac{\frac{1}{2}\times BC\times AM}{\frac{1}{2}\times QR\times PN}=\frac{BC}{QR}\times \frac{AM}{PN}& \\ Now,& \angle B=\angle Q& [\Delta ABC\sim \Delta PQR]\\ & \angle 1=\angle 2& \left[Each{90}^{\circ }\right]\\ \therefore & \Delta ABM\cong \Delta PQN& \left[AAsimilarity\right]\\ \therefore & \frac{AB}{PQ}=\frac{BM}{QN}=\frac{AM}{PN}& \dots \left(ii\right)\\ Also,& \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}& \dots \left(iii\right)[\Delta ABC\sim \Delta PQR]\end{array}$
From (ii) and (iii), $\frac{BC}{QR}=\frac{AM}{PN}$
From(i), $\frac{ar\left(\Delta ABC\right)}{ar\left(\Delta PQR\right)}=\frac{A{M}^2}{P{N}^2}$



$\begin{array}{ccc}\left(ii\right)Prrof:Wehave& \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}& \left[Trianglesaresimilar\right]\dots \left(i\right)\\ \Rightarrow & \frac{AB}{PQ}\Rightarrow \frac{2BD}{2QM}=\frac{AB}{PQ}=\frac{BD}{QM}& \{\begin{array}{c}ADandPMaremedians\\ \therefore BC=2BDandQR=2QM\end{array}\\ and& \angle B=\angle Q& \\ \Rightarrow & \Delta ABD\sim \Delta PQM& \left[AAsimilarity\right]\\ \therefore & \frac{AB}{PQ}=\frac{BL}{QM}=\frac{AL}{PM}& \\ \therefore & \frac{ar\left(\Delta ABC\right)}{ar\left(\Delta PQR\right)}=\frac{A{B}^2}{P{Q}^2}=\frac{A{L}^2}{P{M}^2}& \end{array}$


$\begin{array}{ccc}\left(iii\right)Prrof:Wehave& \angle A=\angle P& [\Delta ABC\sim \Delta PQR]\\ \Rightarrow & \frac{1}{2}\angle A=\frac{1}{2}\angle P& \\ \Rightarrow & \angle 1=\angle 2& \\ and& \angle B=\angle Q& \\ \Rightarrow & \Delta ABL\sim \Delta PQM& \\ \therefore & \frac{AB}{PQ}=\frac{BL}{QM}=\frac{AL}{PM}& \\ \therefore & \frac{ar\left(\Delta ABC\right)}{ar\left(\Delta \right[PQR\left]\right)}=\frac{A{B}^2}{P{Q}^2}=\frac{A{L}^2}{P{M}^2}& \end{array}$