Olympiad Maths Prep

Track / Stage 6 / 17 of 400 #1017 of 2000

Problem 1017

National olympiad, first round
Geometry Difficulty 6.0 Prove it

Let AFBDCEA F B D C E be a convex inscribed hexagon (i.e., whose vertices lie on the same circle) such that

BAD^=DAC^,CBE^=EBA^ and ACF^=FCB^ \widehat{B A D}=\widehat{D A C}, \quad \widehat{C B E}=\widehat{E B A} \quad \text { and } \quad \widehat{A C F}=\widehat{F C B}

Show that the lines (AD)(A D) and (EF)(E F) are perpendicular to each other.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Let PP be the intersection point of (AD)(A D) and (EF)(E F). Then

DPE^=180∘−PED^−EDP^=180∘−(FEB^+BED^)−EDA^=180∘−(FCB^+BAD^)−EBA^=180∘−(12ACB^+12BAC^)−12CBA^=180∘−12(ACB^+BAC^+CBA^)=180∘−12⋅180∘=90∘, \begin{aligned} \widehat{D P E} & =180^{\circ}-\widehat{P E D}-\widehat{E D P} \\ & =180^{\circ}-(\widehat{F E B}+\widehat{B E D})-\widehat{E D A} \\ & =180^{\circ}-(\widehat{F C B}+\widehat{B A D})-\widehat{E B A} \\ & =180^{\circ}-\left(\frac{1}{2} \widehat{A C B}+\frac{1}{2} \widehat{B A C}\right)-\frac{1}{2} \widehat{C B A} \\ & =180^{\circ}-\frac{1}{2}(\widehat{A C B}+\widehat{B A C}+\widehat{C B A}) \\ & =180^{\circ}-\frac{1}{2} \cdot 180^{\circ} \\ & =90^{\circ}, \end{aligned}

which means exactly that (AD)⊥(EF)(A D) \perp(E F).

!

Remark: It is easy to show that D,E,FD, E, F are the midpoints of the arcs BC,CAB C, C A and ABA B.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.