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Belarusian Mathematical Olympiad

Geometry Difficulty 6.2 National Olympiad Prove it Belarus

The rhombus $ABCD$ is given. Let $E$ be one of the points of intersection of the circles $\Gamma_B$ and $\Gamma_C$ , where $\Gamma_B$ is the circle centered at $B$ and passing through $C$ , and $\Gamma_C$ is the circle centered at $C$ and passing through $B$ . The line $ED$ intersects $\Gamma_B$ at point $F$ .
Find the value of the angle $\angle AFB$ .

Solution

Answer: 60°.
We will count the angle $AFB$ as the sum of angles $AFE$ and $BFE$ . Note that $\angle AFE = \angle ACE$ since they share the arc $AE$ in $\Gamma_B$ . And $\angle BFE = \angle BEF$ in the isosceles triangle $BEF$ . The angle $BEF$ equals to the half of the arc $BD$ , which equals to the angle $BCD$ in the circle $\Gamma_C$ , whence $\angle BFE = 0.5\angle BCD$ . The diagonal $AC$ of the rhombus $ABCD$ bisects the angle $DCB$ , hence $\angle BFE = 0.5\angle BCD = \angle ACB$ . Thus

$\angle AFB = \angle AFE + \angle EFB = \angle ECA + \angle ACB = \angle ECB.$
Since the triangle $ECB$ is equilateral, $\angle ECB = 60^\circ$ , and therefore $\angle AFB = 60^\circ$ .

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