Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Austria

We are given an arbitrary acute-angled triangle $ABC$ and its altitudes $AD$ and $BE$ where $D$ and $E$ denote their feet on sides $BC$ and $AC$ , respectively. Let furthermore $F$ and $G$ be two points on segments $AD$ and $BE$ , respectively, such that
$\frac{AF}{FD} = \frac{BG}{GE}.$
The line through $C$ and $F$ intersects $BE$ in point $H$ and the line through $C$ and $G$ intersects $AD$ in point $I$ . Prove that the four points $F, G, H$ and $I$ are concyclic.

Solution

The two right-angled triangles $ADC$ and $BEC$ are inversely similar to each other, see Figure 6. Here, the sides $AD$ and $BE$ correspond to each other.
But the condition
$\frac{AF}{FD} = \frac{BG}{GE}$
means: The two points $F$ and $G$ divide the two sides $AD$ and $BE$ , respectively, in equal ratios. Thus, the two oriented angles $\angle DFC$ and $\angle CGE$ are equal, which implies that the oriented angles $\angle IFH$ and $\angle IGH$ are equal modulo $180^\circ$ . Thus the inscribed angle theorem implies that the four points $F, G, H$ and $I$ are concyclic.

Figure 6: Problem 17

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