Question 14:
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 ∆ABC ~ ∆PQR
Answer:
Given that,
Let us extend AD and PM up to point 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
∴ ∆ABE ∼ ∆PQL (By SSS similarity criterion)
We know that corresponding angles of similar triangles are equal.
∴ ∠BAE = ∠QPL … (1)
Similarly, it can be proved that ∆AEC ∼ ∆PLR and ∠CAE = ∠RPL … (2)
Adding equation (1) and (2), we obtain
∠BAE + ∠CAE = ∠QPL + ∠RPL
⇒ ∠CAB = ∠RPQ … (3)
In ∆ABC and ∆PQR,
AB/PQ = AC/PR (Given)
∠CAB = ∠RPQ [Using equation (3)]
∴ ∆ABC ∼ ∆PQR (By SAS similarity criterion)
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