Question

Question 12
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of triangle PQR. Show that $\Delta ABC\sim \Delta PQR$.

Answer 1

Given,$\Delta ABC$ and $\Delta PQR,AB,BC$ and median AD of $\Delta ABC$ are proportional to sides PQ, QR and median PM of $\Delta PQR$
i.e., $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AD}{PM}$
To Prove that $\Delta ABC\sim \Delta PQR$
Proof
$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AD}{PM}$
$\Rightarrow \frac{AB}{PQ}=\frac{\frac{1}{2}BC}{\frac{1}{2}QR}=\frac{AD}{PM}.....\left(i\right)$
$\Rightarrow \frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$ (D is the mid-point of BC. M is the mid-point of QR)
$\Rightarrow \Delta ABD\sim \Delta PQM$ [SSS similarity criterion]
$\therefore \angle ABD=\angle PQM$ [Corresponding angles of two similar triangles are equal]
$\Rightarrow \angle ABC=\angle PQR$

In $\Delta ABCand\Delta PQR$
$\frac{AB}{PQ}=\frac{BC}{QR}...\left(i\right)$
$\angle ABC=\angle PQR...\left(ii\right)$
From equation (i) and (ii), we get
$\Delta ABC\sim \Delta PQR$ [By SAS similarity criterion]