Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.7 National Olympiad, round 2 Prove it Turkey

In a triangle $ABC$ , the external bisector of $\angle BAC$ intersects the ray $[BC$ at $D$ . The feet of the perpendiculars from $B$ and $C$ to the line $AD$ are $E$ and $F$ respectively, and the foot of the perpendicular from $D$ to $AC$ is $G$ . Show that $\angle DGE + \angle DGF = 180^\circ$ .

Solution

Let $GD$ and $EB$ intersect at the point $P$ . Since $\angle PEA = \angle PGA = 90^\circ$ , the points $P$ , $G$ , $A$ , $E$ are concyclic and hence $\angle EPA = \angle EGA$ . Since $AD$ is the external bisector of $\angle BAC$ , we have $\angle GAD = \angle BAE$ and hence $\angle EBA = \angle ADG$ and it follows that the points $P$ , $D$ , $A$ , $B$ are concyclic. Therefore $\angle BPA = \angle BDA$ . Since the points $C$ , $G$ , $D$ , $F$ are also concyclic, we also get $\angle BDA = \angle FGA$ . Thus, $\angle EGA = \angle FGA$ and hence $\angle DGE + \angle DGF = 90^\circ + \angle EGA + 90^\circ - \angle FGA = 180^\circ$ . Done.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.