Question

In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at a point
Such that:
$PA\times PD=PB\times PC.$

• A. True
• B. False

Answer 1

The correct option is A True

In $△$ APB and $△$ CPD,
$\angle APB=\angle CPD$ (Vertically opposite angles)
$\angle ABP=\angle CDP$ (Alternate angles of parallel sides AB and CD)
$\angle BAP=\angle DCP$ (Alternate angles of parallel sides AB and CD)
Hence, $△APB\sim △CPD$ (AAA rule)
Thus, $\frac{PA}{PC}=\frac{PB}{PD}$
$PA\times PD=PB\times PC$