Question

$\Delta ABC$ and $\Delta PQR$ are two similar triangles shown in the figure. AM and PN are the medians on $\Delta ABC$ and $\Delta PQR$ respectively. Prove that $\Delta ABM$ ~ $\Delta PNQ$ [3 Marks]

Answer 1

Since $\Delta ABC$ and $\Delta PQR$ are similar, ratio of corresponding sides is same.

$\Rightarrow \frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}$

and $\angle A=\angle P,\angle B=\angle Qand\angle C=\angle R$

Medians AM and PN divide BC and QR into two equal parts, respectively.[1 Mark]

$\Rightarrow BM=\frac{BC}{2},QN=\frac{QR}{2}$

In $\Delta ABM$ and $\Delta PQN$,

we know that $\frac{AB}{PQ}=\frac{BC}{QR}$

$\Rightarrow \frac{AB}{PQ}=\frac{2BM}{2QN}$

$\Rightarrow \frac{AB}{PQ}=\frac{BM}{QN}$ [1 Mark]

and $\angle B=\angle Q$

$\therefore \Delta ABM\sim \Delta PQN$ (by SAS similarity criterion) [1 Mark]