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

Answer: Proof that \(\dfrac{A(\Delta ABC)}{A(\Delta PQR)} = \dfrac{AB^2}{PQ^2} = \dfrac{BC^2}{QR^2} = \dfrac{AC^2}{PR^2}\)

[Figure: \(\Delta ABC\) with altitude AD on BC; \(\Delta PQR\) with altitude PS on QR.]

Given: \(\Delta ABC \sim \Delta PQR\), so \(\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR}\) and \(\angle B \cong \angle Q\).

Construction: Draw seg AD \(\perp\) BC and seg PS \(\perp\) QR.

Proof: \(\dfrac{A(\Delta ABC)}{A(\Delta PQR)} = \dfrac{\frac{1}{2} BC \times AD}{\frac{1}{2} QR \times PS} = \dfrac{BC}{QR} \times \dfrac{AD}{PS}\) ...... (I)

In \(\Delta ABD\) and \(\Delta PQS\): \(\angle B \cong \angle Q\) (given) and \(\angle ADB \cong \angle PSQ\) (each 90°). \(\therefore \Delta ABD \sim \Delta PQS\) (AA test) \(\therefore \dfrac{AD}{PS} = \dfrac{AB}{PQ}\).

But \(\dfrac{AB}{PQ} = \dfrac{BC}{QR}\), so \(\dfrac{AD}{PS} = \dfrac{BC}{QR}\) ...... (II)

From (I) and (II), \(\dfrac{A(\Delta ABC)}{A(\Delta PQR)} = \dfrac{BC}{QR} \times \dfrac{BC}{QR} = \dfrac{BC^2}{QR^2}\).

Since \(\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR}\), this also equals \(\dfrac{AB^2}{PQ^2}\) and \(\dfrac{AC^2}{PR^2}\). Hence proved.