Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.9 AIME, harder Prove it

In a triangle, $\angle A=45^{\circ}, \angle B=75^{\circ}$ . The altitude from $A$ meets $BC$ at $D$ , and the altitude from $C$ meets $AB$ at $F$ . The perpendicular from $F$ to $AC$ meets $AC$ at $G$ . Prove that $AG=GD$ .

Solution

Since the triangle is acute-angled, both altitudes intersect the opposite sides at internal points. The triangle $AFC$ is right-angled and one of its angles is $45^{\circ}$ , so the other angle, $\angle ACF$ , is also $45^{\circ}$ . Therefore, $FG$ bisects the base $AC$ of the triangle.

# $1984-05-216-2 . e p s$

The Thales circle drawn over the segment $AC$ as a diameter passes through the points $F$ and $D$ . The center of the circle, which is $G$ , is equidistant from $A$ , $F$ , $D$ , and $C$ , i.e., $AG = GD$ , which is what we wanted to prove.

Remark. We did not use the fact that $\angle B = 75^{\circ}$ , so the statement of the problem is true for any triangle where one of the angles is $45^{\circ}$ .

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