Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.8 National olympiad Prove it Croatia

Let $\overline{BD}$ and $\overline{CE}$ be the altitudes of an acute-angled triangle $ABC$ . The circle with diameter $\overline{AC}$ meets $\overline{BD}$ at $F$ . The circle with diameter $\overline{AB}$ meets the line $CE$ at points $G$ and $H$ , where $G$ is between $C$ and $E$ . If $\angle CHF = 12^\circ$ , find the measure of the angle $\angle AGF$ . (Go Geometry)

Solution

The chord $\overline{GH}$ is perpendicular to $\overline{AB}$ , so $AB$ is the bisector of the segment $\overline{GH}$ . Hence $|AG| = |AH|$ .

Since $AFC$ is a right-angled triangle, Euclid's theorem gives us $|AF|^2 = |AD| \cdot |AC|$ .

Analogously, since $ABG$ is a right-angled triangle, we have $|AG|^2 = |AE| \cdot |AB|$ .

Angles $\angle BDC$ and $\angle BEC$ are right angles, so the quadrilateral $BCDE$ is cyclic. By using the power-of-a-point theorem applied to $A$ with respect to the circle circumscribed to the quadrilateral $BCDE$ , we conclude that $|AD| \cdot |AC| = |AE| \cdot |AB|$ .

Hence $|AF| = |AG| = |AH|$ , i.e. the point $A$ is the circumcentre of the triangle $GFH$ .

Finally,
$\angle AGF = \angle GFA = \frac{1}{2}(180^\circ - \angle FAG) = \frac{1}{2}(180^\circ - 2\angle FHG) = \frac{1}{2}(180^\circ - 24^\circ) = 78^\circ.$

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