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Ukrainian National Mathematical Olympiad

Geometry Difficulty 5.1 AIME, harder Prove it Ukraine

Let $ABCD$ be a parallelogram. The circle through $A$ and $D$ intersects the lines $AB$ , $BD$ , $AC$ and $CD$ in points $B_1$ , $B_2$ , $C_1$ and $C_2$ respectively. Let $K$ be the intersection point of the lines $B_1B_2$ and $C_1C_2$ . Prove that $K$ is equidistant from the lines $AB$ and $CD$ .

Fig. 41

Solution

Let $O$ be the intersection point of the diagonals of the parallelogram. Let's prove that the points $C_1$ , $O$ , $B_2$ and $K$ lie on the same circle. In fact, with the cyclicity of points $A$ , $B_1$ , $B_2$ , $C_1$ , $C_2$ and $D$ and parallelism of $AB$ and $CD$ we have (Fig. 41):
$\angle(KC_1, C_1O) = \angle(CD, DA) = \angle(BA, AD) = \angle(B_1B_2, B_2D) = \angle(B_2K, B_2O).$
Now $\angle(BO, OK) = \angle(B_2C_1, C_1K) = \angle(B_2D, DC)$ , thus $OK \parallel DC$ . As known, diagonals $AC$ and $BD$ are divided by a point $O$ in half. By the Thales' theorem the line $OK$ (and accordingly, the point $K$ ) is equidistant from the lines $AB$ and $CD$ .

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