Question

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC∼ΔPQR

Answer 1

Given: $\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$

To Prove: ΔABC∼ΔPQR

Proof: Let us extend AD and PM up to points E and L respectively, such that AD = DE and PM = ML.

Then, join B to E, C to E, Q to L, and R to L.

We know that medians divide opposite sides.
Therefore, BD = DC and QM = MR

Also, AD = DE (By construction)
And, PM = ML (By construction)

In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.
Therefore, quadrilateral ABEC is a parallelogram.
∴ AC = BE and AB = EC (Opposite sides of a parallelogram are equal)

Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR

It was given that
$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$
⇒$\frac{AB}{PQ}=\frac{BE}{QL}=\frac{2AD}{2PM}$
⇒$\frac{AB}{PQ}=\frac{BE}{QL}=\frac{AE}{PL}$
∴ ∆ABE ∼ ∆PQL (By SSS similarity criterion)

We know that the corresponding angles of similar triangles are equal.
∴ ∠BAE = ∠QPL ……… (1)

Similarly, it can be proved that ∆AEC ∼ ∆PLR and,
$\angle$CAE = $\angle$RPL ……… (2)

Adding equation (1) and (2), we obtain
$\angle$BAE + $\angle$CAE = $\angle$QPL + $\angle$RPL
⇒ $\angle$CAB = $\angle$RPQ ………. (3)

In ∆ABC and ∆PQR,
$\frac{AB}{PQ}=\frac{AC}{PR}$(Given)
$\angle$CAB =$\angle$RPQ [Using equation (3)]
∴ ∆ABC ∼ ∆PQR (By SAS similarity criterion)