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

We need to prove that the ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding sides. Let the triangles be ∆𝐴𝐵𝐶 and ∆𝑃𝑄R.

Now, according to the question, ΔABC ~ ΔPQR and we need to prove that

$$\frac{ar\left(𝛥ABC\right)}{ar\left(𝛥PQR\right)}={\left(\frac{AB}{PQ}\right)}^2$$

On drawing AD⟂BC and PE⟂QR, we get the diagram as shown below

Since ΔABC ~ ΔPQR, it can be said that the corresponding angles are equal. 

Therefore, both the triangles are equiangular and hence their corresponding sides are proportional. It can be written as $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR} ...\left(i\right)$

Also, since

ΔABC ~ ΔPQR, it can be said thatΔADB ~ ΔPEQ.  Therefore, 

$\frac{AD}{PE}=\frac{AB}{PQ} .......\left(ii\right)$

Now, from the above two equations, 

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=\frac{AD}{PE} ......\left(iii\right)$

Now, the ratio of areas of ΔABC  and ΔPQR  is given as 

$$\frac{ar\left(𝛥ABC\right)}{ar\left(𝛥PQR\right)}=\frac{0.5\times BC\times AD}{0.5\times QR\times PE}$$$$\Rightarrow \frac{ar\left(𝛥ABC\right)}{ar\left(𝛥PQR\right)}=\frac{BC}{QR}\times \frac{AD}{PE}$$

From the third equation, we get 

$$\Rightarrow \frac{ar\left(𝛥ABC\right)}{ar\left(𝛥PQR\right)}={\left(\frac{AB}{PQ}\right)}^2$$

Hence, proved.