Saudi Arabia Mathematical Competitions 2012

Since $AED$ and $ADC$ are both right triangles, $\widehat{ADP}=90^{\circ }-\widehat{DAC}=\widehat{ECB}$. Also, since $AFDB$ is a cyclic quad, $\widehat{DAP}=\widehat{EBC}$.



Therefore, $ADP$ and $BCE$ are similar and $\frac{BC}{EC}=\frac{AD}{DP}$. Also, $ADC$ and $DEC$ are similar since both are right angled triangles that share an acute angle. Therefore, $$\frac{AD}{DC}=\frac{DE}{EC}.$$ These two equations imply that $$BC\cdot DP=AD\cdot EC=DC\cdot DE,$$ so we have $$\frac{BC}{DC}=\frac{DE}{DP}.$$ Since $BC=BD+DC$ and $DE=DP+PE$, subtracting one from both sides gives us $\frac{BD}{DC}=\frac{PE}{DP}$, as desired.