AUT_ABooklet_2020

Question

We are given an arbitrary acute-angled triangle

A B C

and its altitudes

A D

and

B E

where

D

and

E

denote their feet on sides

B C

and

A C

, respectively. Let furthermore

F

and

G

be two points on segments

A D

and

B E

, respectively, such that

\frac{A F}{F D} = \frac{B G}{GE} .

The line through

C

and

F

intersects

B E

in point

H

and the line through

C

and

G

intersects

A D

in point

I

. Prove that the four points

F, G, H

and

I

are concyclic.

Accepted Answer

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 AFFD = BGGE means: The two points F and G divide the two sides AD and BE, respectively, in equal ratios. Thus, the two oriented angles DFC and CGE are equal, which implies that the oriented angles IFH and IGH are equal modulo 180^. Thus the inscribed angle theorem implies that the four points F, G, H and I are concyclic. FIGURE_0 Figure 6: Problem 17