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

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Step 1: Given: $	riangle ABC \sim 	riangle PQR$. To Prove: $\frac{	ext{ar}(	riangle ABC)}{	ext{ar}(	riangle PQR)} = \frac{AB^2}{PQ^2} = \frac{BC^2}{QR^2} = \frac{AC^2}{PR^2}$. Construction: Draw $AM \perp BC$ and $PN \perp QR$.
Step 2: $\frac{	ext{ar}(	riangle ABC)}{	ext{ar}(	riangle PQR)} = \frac{\frac{1}{2} 	imes BC 	imes AM}{\frac{1}{2} 	imes QR 	imes PN} = \frac{BC 	imes AM}{QR 	imes PN}$
Step 3: In $	riangle ABM$ and $	riangle PQN$: $\angle B = \angle Q$ (similar triangles), $\angle AMB = \angle PNQ = 90°$. So $	riangle ABM \sim 	riangle PQN \Rightarrow \frac{AM}{PN} = \frac{AB}{PQ} = \frac{BC}{QR}$.
Step 4: $	herefore \frac{	ext{ar}(	riangle ABC)}{	ext{ar}(	riangle PQR)} = \frac{BC}{QR} 	imes \frac{AM}{PN} = \frac{BC}{QR} 	imes \frac{BC}{QR} = \frac{BC^2}{QR^2}$. Similarly $= \frac{AB^2}{PQ^2} = \frac{AC^2}{PR^2}$.

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