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Saudi Arabian Mathematical Competitions, 2012

Geometry Difficulty 4.7 AIME Prove it Saudi Arabia

In triangle $ABC$ , points $D$ and $E$ lie on side $BC$ and $AC$ respectively such that $AD \perp BC$ and $DE \perp AC$ . The circumcircle of triangle $ABD$ meets segment $BE$ at point $F$ (other than $B$ ). Ray $AF$ meets segment $DE$ at point $P$ . Prove that
$\frac{DP}{PE} = \frac{CD}{DB}.$

Solution

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.

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