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)