The ratio of the area of …

Given : {tex}\Delta A B C \sim \Delta P Q R{/tex}
To Prove : {tex}\frac { \operatorname { ar } ( \Delta A B C ) } { \operatorname { ar } ( \Delta P Q R ) } = \left( \frac { A B } { P Q } \right) ^ { 2 }= \left( \frac { BC } { QR } \right) ^ { 2 }= \left( \frac { A C } { P R } \right) ^ { 2 }{/tex}
Construction: Draw AD {tex} \bot {/tex}BC and PE {tex} \bot {/tex} QR
Proof :

{tex}\Delta A B C \sim \Delta P Q R{/tex} 
{tex}\therefore\frac { A B } { P Q } = \frac { B C } { Q R } = \frac { A C } { P R }{/tex} ( Ratio of corresponding sides of similar triangles are equal) ...(i)
{tex}\angle B = \angle Q{/tex} (Corresponding angles of similar triangles)......... (ii)
In {tex}\Delta A D B \text { and } \Delta P E Q{/tex} 
{tex}\angle B = \angle Q{/tex} ( From (ii)) 
{tex}\angle A D B = \angle P E Q{/tex} {tex}\left[\; \operatorname { each } 90 ^ { \circ } \right]{/tex} 
{tex}\therefore {/tex} {tex}\Delta A D B \sim \Delta P E Q{/tex} [ By AA criteria] 
{tex}\Rightarrow{/tex} {tex}\frac { A D } { P E } = \frac { A B } { P Q }{/tex}  (Corresponding sides of similar triangles) ...(iii)
From equation (i) and equation (iii)
{tex}\frac { A B } { P Q } = \frac { B C } { Q R } = \frac { A C } { P R } = \frac { A D } { P E }{/tex} ...(iv)
{tex}\frac { \operatorname { ar } ( \Delta A B C ) } { \operatorname { ar } ( \Delta P Q R ) } = \frac { \frac { 1 } { 2 } \times B C \times A D } { \frac { 1 } { 2 } \times Q R \times P E }{/tex} 
{tex}= \left( \frac { B C } { Q R } \right) \times \left( \frac { A D } { P E } \right){/tex} 
({tex}\frac{AD}{PE}=\frac{BC}{QR}{/tex}) 
{tex}= \frac { B C } { Q R } \times \frac { B C } { Q R }{/tex} 
{tex}\Rightarrow{/tex} {tex}\frac { a r ( \Delta A B C ) } { a r ( \Delta P Q R ) } = \frac { B C ^ { 2 } } { Q R ^ { 2 } }{/tex} ....(v) [from eq. (iv)]
From equation (iv) and equation (v),
{tex}\therefore\frac { \operatorname { ar } ( \Delta A B C ) } { \operatorname { ar } ( \Delta P Q R ) } = \left( \frac { A B } { P Q } \right) ^ { 2 }= \left( \frac { BC } { QR } \right) ^ { 2 }= \left( \frac { A C } { P R } \right) ^ { 2 }{/tex}
{tex}\therefore{/tex} Ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.