In the following figure, line \(l\) intersects side AB and side AC of \(\Delta ABC\) in points P and Q respectively. Show that \(\dfrac{A(\Delta APQ)}{A(\Delta ABC)} = \dfrac{AP \times AQ}{AB \times AC}\).[Figure: \(\Delta ABC\) with A at the top, B…

Answer: Proof by comparing areas of triangles with a common vertex.

Given: line \(l\) intersects side AB at P and side AC at Q in \(\Delta ABC\).

To prove: \(\dfrac{A(\Delta APQ)}{A(\Delta ABC)} = \dfrac{AP \times AQ}{AB \times AC}\).

Proof: \(\Delta APQ\) and \(\Delta ABQ\) have the same height from Q to line AB, and their bases are AP and AB on the same line.

\(\therefore \dfrac{A(\Delta APQ)}{A(\Delta ABQ)} = \dfrac{AP}{AB}\) .............. (I)

Similarly, \(\Delta ABQ\) and \(\Delta ABC\) have the same height from B to line AC, and their bases are AQ and AC.

\(\therefore \dfrac{A(\Delta ABQ)}{A(\Delta ABC)} = \dfrac{AQ}{AC}\) .............. (II)

Multiplying (I) and (II),

\(\dfrac{A(\Delta APQ)}{A(\Delta ABQ)} \times \dfrac{A(\Delta ABQ)}{A(\Delta ABC)} = \dfrac{AP}{AB} \times \dfrac{AQ}{AC}\)

\(\therefore \dfrac{A(\Delta APQ)}{A(\Delta ABC)} = \dfrac{AP \times AQ}{AB \times AC}\), as required.