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

We have

Given: ∆ABC ∼ ∆PQR

To prove: Ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

i.e.,

Construction: Draw AM perpendicular to BC and PN perpendicular to QR.

Proof: We know area of the triangle is given by (1/2 × base × height).

So in ∆ABC,

Area (∆ABC) = 1/2 × BC × AM [∵ base = BC & height = AM] …(i)

Similarly, in ∆PQR

Area (∆PQR) = 1/2 × QR × PN [∵ base = QR & height = PN] …(ii)

Dividing equations (i) by (ii), we get

⇒ …(iii)

In ∆ABM & ∆PQN,

∠B = ∠Q [∵ ∆ABC ∼ ∆PQR; and corresponding angles are equal of similar triangles]

∠AMB = ∠PNQ [∵ they are 90°]

So by AA-similarity property of triangle, ∆ABM ∼ ∆PQN.

⇒ [∵ corresponding sides of similar triangles are proportional] …(iv)

Substituting equation (iv) in equation (iii), we have

⇒ …(v)

We know that,

∆ABC ∼ ∆PQR

Using property which says that corresponding sides of similar triangles are proportional, we can write as

Using this equality, re-write equation (v)

⇒

⇒ …(A)

Using this equality again, re-writing equation (v)

⇒

⇒ …(B)

Similarly, …(C)

Collecting equations (A), (B) & (C), we get

Hence, proved.