Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

### Answer

Given: Δ ABC ~ Δ PQR

To Prove: ar(ΔABC) / ar(ΔPQR) = (AB/PQ)^{2} = (BC/QR)^{2} = (CA/RP)^{2}

Construction: Draw AM ⊥ BC, PN ⊥ QR

ar(ΔABC) / ar(ΔPQR) = (½ × BC × AM) / (½ × QR × PN)

= BC/QR × AM/PN ... [i]

In Δ ABM and Δ PQN,

∠B = ∠Q (Δ ABC ~ Δ PQR)

∠M = ∠N (each 90°)

So, Δ ABM ~ Δ PQN (AA similarity criterion)

Therefore, AM/PN = AB/PQ ... [ii]

But, AB/PQ = BC/QR = CA/RP (Δ ABC ~ Δ PQR) ... [iii]

Hence, from (i)

ar(ΔABC) / ar(ΔPQR) = BC/QR × AM/PN

= AB/PQ × AB/PQ [From (ii) and (iii)]

= (AB/PQ)^{2}

Using (iii)

ar(ΔABC) / ar(ΔPQR) = (AB/PQ)^{2} = (BC/QR)^{2} = (CA/RP)^{2}