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Belarus — Selection and Training Session

Geometry Difficulty 6.7 National Olympiad Prove it Belarus

Let $\Omega$ be a circle, $S$ be a point outside $\Omega$ and $SA, SB$ be tangents from $S$ to $\Omega$ . Arbitrary point $K$ is chosen on the segment $AB$ . Points $U, W, R$ and $T$ are chosen on $\Omega$ such that $UW \cap RT = K$ and the points $W, T$ lie on the same half-plane with respect to the line $AB$ . The line $SK$ intersects $\Omega$ at $P$ and $Q$ and $P$ lies on the same half-plane to the line $AB$ as $W$ and $T$ . The lines $WR$ and $TU$ intersect $AB$ at $C$ and $D$ respectively. Let $M$ be the midpoint of $PQ$ .
Prove that $\angle AMC = \angle BMD$ .

Solution

We will use multiple times the following well-known
Lemma. Let the chords $AB$ and $CD$ of a circle intersect at the point $K$ then
$\frac{AC \cdot AD}{BC \cdot BD} = \frac{CK}{DK}$
Proof of lemma. This equality directly follows from the sine laws for the triangles $ACK, ADK$ and $BCD$ .

The quadrilateral $APBQ$ is harmonic, so $MK$ is the bisector of the angle $AMB$ . Hence $\frac{AC}{CB} = \frac{\sin AMC}{\sin CMB} \cdot \frac{AK}{BK}$ and $\frac{AD}{DB} = \frac{\sin AMD}{\sin DMB} \cdot \frac{AK}{BK}$ , therefore it's enough to prove the equality $\frac{AC}{CB} \cdot \frac{AD}{DB} = \left(\frac{AK}{BK}\right)^2$ .

The lemma for $AB \cap RW = C$ and $AB \cap UT = D$ implies that
$\frac{AC}{CB} = \frac{AW \cdot AR}{BW \cdot BR} \quad \text{and} \quad \frac{AD}{DB} = \frac{AT \cdot AU}{BT \cdot BU}$
And the lemma for $AB \cap RT = K$ and $AB \cap UW = D$ implies that
$\frac{AK}{KB} = \frac{AT \cdot AR}{BT \cdot BR} = \frac{AW \cdot AU}{BW \cdot BU}$
Combining these equalities we obtain the required equality.

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