Let $ABC$ be an acute-angled triangle. Let $D$ and $E$ be the feet of the altitudes on the sides $BC$ or $AC$. Points $F$ and $G$ are located on the lines $AD$ and $BE$ in such a way that$ \frac{AF}{FD}=\frac{BG}{GE}$. The line passing through $C$ and $F$ intersects $BE$ at point $H$, and the line passing through $C$ and $G$ intersects $AD$ at point $I$. Prove that points $F, G, H$ and $I$ lie on a circle. (Walther Janous)
Problem
Source: 2019 Austrian Federal Competition For Advanced Students, Part 2 p5
Tags: ratio, equal ratio, Concyclic, geometry
MP8148
06.03.2020 10:11
[asy][asy]
size(7cm);
defaultpen(fontsize(10pt));
pair A = dir(120), B = dir(210), C = dir(330), D = foot(A,B,C), E = foot(B,C,A), F = A+dir(A--D)*abs(A-D)*0.85, G = B+dir(B--E)*abs(B-E)*0.85, H = extension(C,F,B,E), I = extension(C,G,A,D);
dot("$A$", A, dir(120));
dot("$B$", B, dir(210));
dot("$C$", C, dir(330));
dot("$D$", D, dir(270));
dot("$E$", E, dir(45));
dot("$F$", F, dir(50));
dot("$G$", G, dir(90));
dot("$H$", H, dir(90));
dot("$I$", I, dir(45));
draw(A--B--C--A);
draw(A--D^^B--E);
draw(H--C--I);
draw(circumcircle(H,F,G),dotted);
[/asy][/asy]
By the algebraic identity $$k = \dfrac ab = \dfrac cd \implies k = \dfrac{a+c}{b+d},$$we have $$\dfrac{DF}{EG} = \dfrac{DF+FA}{EG+GB} = \dfrac{AD}{BE}.$$Since $\triangle ADC \sim \triangle BEC$, we know that $$\dfrac{AD}{BE} = \dfrac{AC}{BC} = \dfrac{CD}{CE} \implies \dfrac{DF}{EG} = \dfrac{CD}{CE},$$so $\triangle CDF \sim \triangle CEG$ by SAS. Then $$\angle GHF = \angle HBC + \angle HCB = \angle IAC + \angle ICA = \angle GIF$$as desired.
Pluto04
06.03.2020 12:18
Maybe a similar solution but still I will post.
Solution:
$ \frac{AF}{FD}+1=\frac{BG}{GE}+1$
$\implies \frac{AD}{FD}=\frac{BE}{GE}\implies \frac{DF}{GE}=\frac{AD}{BE}$
Moreover,
$\angle ACD = \angle BCE.$
$\angle ADC= \angle BEC.$
Hence by AA criterion,
$\triangle ADC \sim \triangle BEC$
$\implies \frac{DF}{EG} = \frac{CD}{CE}$
$\implies \triangle CDF \sim \triangle CEG$
$\angle GHF = \angle HBC + \angle HCB = \angle IAC + \angle ICA = \angle GIF$
Hence, G,H,I,F are concyclic.
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