, 2018

Geometry Difficulty 8.1 Shortlist Prove it Saudi Arabia

Let $ABCD$ be a convex quadrilateral inscribed in circle $(O)$ such that $DB = DA + DC$ . The point $P$ lies on the ray $AC$ such that $AP = BC$ . The point $E$ is on $(O)$ such that $BE \perp AD$ . Prove that $DP$ is parallel to the angle bisector of $\angle BEC$ .

Solution

Denote $R, F$ as the intersections of $AC$ with the angle bisector of $\angle BEC$ and $BE$ , respectively. Let $Q$ be a point on $BD$ such that $DQ = DC$ , hence $QB = DB - DQ = DB - DC = AD$ .

Thus two triangles $CQB$ and $PDA$ are congruent (s.a.s). Note that $AD \perp BE$ then $\angle CAD + \angle AFE = 90^\circ$ . From these, we have

$\begin{aligned} \angle APD & = \angle BCQ = \angle CQD - \angle CBD \\ & = 90^\circ - \frac{1}{2} \angle BDC - \angle CAD = \left(90^\circ - \angle CAD\right) - \frac{1}{2} \angle BEC \\ & = \angle AFE - \angle BER = \angle ARE . \end{aligned}$

Hence, we conclude that $DP \parallel ER$ . $\square$

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.