NCERT Solutions
Class 10 Maths
Triangles

Ex. 6.3 Q.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: Δ ABC ~ Δ PQR.
Given that,
AB/PQ = AC/PR = AD/PM
Produce AD and PM to E and L such that AD = DE and PM = DE
Now, join B to E, C to E, Q to L and R to L.
AD and PM are medians of triangles, therefore
BD and Dc and QM = MR
AD = DE and PM = ML [By construction]
So, the diagonals of ABEC bisecting each other at D,
Therefore, ABEC is a parallelogram.
So, AC = BE, AB = EC, PR = QL and PQ = LR
Given that, AB/PQ = AC/PR = AD/PM
=> AB/PQ = BE/QL = 2AD/2PM

=> AB/PQ = BE/QL = AE/PL
Hence, Δ ABE ~ Δ PQL [SSS similarity]
We know that the corresponding angles of similar triangles are equal.
Therefore, ∠ BAE = ∠ QPL ……….1
Similarly, Δ AEC ~ Δ PLR
and ∠ CAE = ∠ RPL ……….2
Add equation 1 and 2, we get
∠ BAE + ∠ CAE = ∠ QPL + ∠ RPL
=> ∠ CAB = ∠ RPQ ……….3
In Δ ABC and Δ PQR,
AB/PQ = AC/PR [Given]
∠ CAB = ∠ RPQ [Proved]
So, Δ ABC ~ Δ PQR [SAS similarity]