Question

In $\square ABCD$, side $BC\parallel$ side $AD$. Diagonals $AC$ and $BD$ intersect each other at $P$. If $AP=\frac{1}{3}AC$, then prove that $DP=\frac{1}{2}BP$.

Answer 1

Clearly, $△APD\sim △CPB$

$\Rightarrow \frac{AP}{CP}=\frac{PD}{PB}$

$\Rightarrow \frac{AP}{AC-AP}=\frac{PD}{PB}$

$\Rightarrow \frac{1}{\frac{AC}{AP}-1}=\frac{PD}{PB}$

$\Rightarrow \frac{1}{3-1}=\frac{PD}{PB}$

$\therefore DP=\frac{1}{2}BP$