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