Competition problem — Geometry — Olympiad Maths Prep

Geometry Difficulty 7.0 National olympiad, round 2 Prove it Belarus

$ABCD$ is a cyclic quadrilateral. Let the circle $\Gamma_1$ pass through points $A$ and $B$ and touch $CD$ at point $E$ ; let the circle $\Gamma_2$ pass through points $B$ and $C$ and touch $DA$ at point $F$ ; let the circle $\Gamma_3$ pass through points $C$ and $D$ , and touch $AB$ at point $G$ ; let at last the circle $\Gamma_4$ pass through points $D$ and $A$ and touch $BC$ at point $H$ .
Prove that $EG \perp FH$ .
(A. Voidelevich)

Solution

If the quadrilateral $ABCD$ has the pair of parallel sides the result follows from the symmetry of the construction. So we suppose that $ABCD$ is different from a trapezoid and a rectangle.
Let $X = BC \cap AD$ . Since $\Gamma_2$ touches $AD$ at point $F$ , we have $XF^2 = XB \cdot XC$ .

Similarly, $XH^2 = XA \cdot XD$ . Since $ABCD$ is cyclic, we have $XB \cdot XC = XA \cdot XD$ , so $XF = XH$ , and hence $\triangle XHF$ is an isosceles triangle. Thus $\angle BHF = 0.5(180^\circ - \angle CXD) = 0.5(\angle C + \angle D)$ . Similarly, $\angle BGE = 0.5(\angle A + \angle D)$ . Let $GE \cap HF = Y$ . Then $\triangle GYH = 360^\circ - (\angle B + 0.5(\angle C + \angle D)) + 0.5(\angle A + \angle D)) = 360^\circ - (\angle B + \angle D) - 0.5(\angle A + \angle C)$ . Since $\angle A + \angle C = \angle B + \angle D = 180^\circ$ , we have $\triangle GYH = 360^\circ - 180^\circ - 90^\circ = 90^\circ$ , so $EG \perp FH$ .

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