Question

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR (see the given figure). Show that ∆ABC ∼ ∆PQR.

Answer 1

Given: Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR
$\Rightarrow \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AD}{PM}$

To Prove: ∆ABC ∼ ∆PQR

Proof: The median divides the opposite side.
∴ BD=$\frac{BC}{2}$ and QM=$\frac{QR}{2}$

Given that,

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AD}{PM}$

⇒ $\frac{AB}{PQ}=\frac{\frac{1}{2}BC}{\frac{1}{2}QR}=\frac{AD}{PM}$

⇒ $\frac{AP}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$

In ∆ABD and ∆PQM,
$\frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$
∴ ∆ABD ∼ ∆PQM (By SSS similarity criterion)
⇒ $\angle$ABD = $\angle$PQM (Corresponding angles of similar triangles)

In ∆ABC and ∆PQR,
⇒ $\angle$ABD = $\angle$PQM (Proved above)
⇒ $\frac{AB}{PQ}=\frac{BC}{QR}$
∴ ∆ABC ∼ ∆PQR (By SAS similarity criterion)

Hence Proved