If a line parallel to $B C$ intersects side $A B$ and $A C$ of triangle $A B C$ at $P$ and $Q,$ then prove that $\frac{A P}{A B} = \frac{A Q}{A C} .$

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Solution

In $△ A P Q$ and $△ A B C$ ,

$∠ A$ is common.

$∠ A P Q = ∠ A B C$ ...[corresponding angles]

$∠ A Q P = ∠ A C B$ ...[corresponding angles]

$∴ △ A P Q \sim △ A B C$ ...(By $A A A$ )

So, once the triangles are similar, their corresponding sides would be proportional.

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

Hence proved.

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