Explanation by Vanshika
Solution :
Given : AB/PQ = AC/PR = AD/PM
To Prove : Δ ABC ~ Δ PQR
Construction : Draw DE||AB and MF||PQ
Proof : DE||AB (By Construction)
D is the mid point of BC (AD is a Median)
⇒ E is the mid point of AC (By Converse of Mid Point Theorem)
⇒ DE=½AB — (1)
Similarly, F is the mid point of PR
MF=½PQ — (2)
Dividing (1) by (2)
DE/MF = AB/PQ — (3)
AB/PQ = AC/PR = AD/PM (Given)
DE/MF = 2AE/2PF = AD/PM [Using (1), (2) and (3)]
DE/MF = AE/PF = AD/PM
In ΔADE and ΔPMF
DE/MF = AE/PF = AD/PM (Proved above)
ΔADE~ΔPMF (By SSS)
∠1 = ∠2 (By CPST) — (4)
similarly,
∠3 = ∠4 — (5)
Adding equation (4) and (5)
∠1+∠3=∠2+∠4
∠A=∠P
Now, In ΔABC and ΔPQR
AB/PQ=AC/PR (Given)
∠A=∠P (Proved above)
∴ Δ ABC ~ Δ PQR (By SAS)
Hence Proved