Let $A B C$ be a triangle and $D$ and $E$ be two points on side $A B$ such that $A D = B E$ . If $D P | | B C$ and $E Q | | A C$ , then prove that $P Q | | A B$ .

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Solution

In $△ A B C$ , we have

$D P | | B C$ and $E Q | | A C$

$∴$ $\frac{A D}{D B} = \frac{A P}{P C}$ and $\frac{B E}{E A} = \frac{B Q}{Q C}$

$\Rightarrow$ $\frac{A D}{D B} = \frac{A P}{P C}$ and $\frac{A D}{D B} = \frac{B Q}{Q C}$

$\Rightarrow$ $\frac{A P}{P C} = \frac{B Q}{Q C}$

$\Rightarrow$ In a $△ A B C$ , P and Q divide sides CA and CB respectively in the same ratio.

$∴$ $P Q | | A B$ [By the converse of Basic Proportionality Theorem]

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