Competition problem — Geometry — Maths Olympiad Prep

Library / /318 of 520

Geometry Difficulty 6.7 National olympiad Prove it

Let $ABC$ be an isosceles triangle at $A$ , and let $D$ be a point on ( $AC$ ) such that $A$ is located between $C$ and $D$ , but is not the midpoint of [CD].
We denote $d_{1}$ and $d_{2}$ as the internal and external bisectors of the angle $\widehat{BAC}$ , and $\Delta$ as the perpendicular bisector of [BD]. Finally, let $E$ and $F$ be the points of intersection of $\Delta$ with the lines $d_{1}$ and $d_{2}$ , respectively.
Prove that the points $A, D, E$ , and $F$ are concyclic.

Solution

A pretty figure suggests that points $A, B, D, E$ , and $F$ are concyclic: this is what we will show.
Let $\Gamma$ be the circumcircle of $A B D$ , and let $G_{1}$ and $G_{2}$ be the points of intersection of lines $d_{1}$ and $d_{2}$ with $\Gamma$ , and other than $D$ itself. It suffices to show that $E=G_{1}$ and that $\mathrm{F}=\mathrm{G}_{2}$ .
Since $\mathrm{d}_{2}$ is the external bisector of $\widehat{B A C}$ , it is the internal bisector of $\widehat{B A D}$ , and the South Pole theorem indicates that $\mathrm{G}_{2}$ is equidistant from B and D. Similarly, $\mathrm{d}_{1}$ is the internal bisector of $\widehat{B A C}$ , hence the external bisector of $\widehat{B A D}$ , and the North Pole theorem indicates that $\mathrm{G}_{1}$ is equidistant from B and D.
But then $\left(G_{1} G_{2}\right)$ is indeed the perpendicular bisector of $[B D]$ , so $G_{1}=E$ and $G_{2}=F$ , which concludes.
!

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