Proof: We are given two triangles ABC and PQR such that Δ ABC ~Δ PQR
We need to prove that ar(ABC) / ar(PQR) = (AB/PQ)² = (BC/QR)²=(CA/RP)²
For Finding the areas of the two triangles, we draw altiude AM and PN of the triangle
Now, ar(ABC) = 1/2 BC x AM and ar(PQR) = 1/2 QR x PN
Therefore , AM/PN = AB/PQ
Also, Δ ABC ~ Δ PQR
So, AB/PQ = BC/QR = CA/RP
Therefore, ar(ABC)/ar(PQR) = AB/PQ x AM/PN = AB/PQ x AB/PQ = (AB/PQ)²
Weget ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)²
So, ar(ABC)/ar(PQR) = (1/2 x BC x AM) / 1/2 x QR x PN = (BC x AM) / (QR x PN)
Now, in Δ ABM and Δ PQN, ∠B = ∠Q (As Δ ABC ~ Δ PQR)
∠M = ∠N (Each is of 90⁰)
and Δ ABM ~Δ PQN (AA similarity criterion)