Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.1 National olympiad Prove it Croatia

Let $p$ and $q$ be two parallel lines. Circle $k$ touches the line $p$ at $A$ and intersects $q$ at two different points, $B$ and $C$ . Let $T$ be some point on $p$ . Segments $\overline{TB}$ and $\overline{TC}$ intersect the shorter arc $\widehat{AC}$ at $K$ and $L$ respectively. Points $K$ and $L$ are both different from $B$ and $C$ .
Prove that the line $KL$ passes through the midpoint of the segment $\overline{AT}$ .

Solution

Let $P$ be the intersection of the lines $p$ and $KL$ . Denote $\angle TBC = x$ .

Line $BT$ is a transversal of the parallel lines $p$ and $q$ , which implies that $\angle BTA = \angle TBC = x$ .
The quadrilateral $BCLK$ is cyclic, hence $\angle CLK = 180^\circ - \angle KBC = 180^\circ - x$ , implying $\angle KLT = x$ .
Triangles $PKT$ and $PTL$ are similar because they share an angle at $P$ and $\angle KTP = \angle PLT = x$ . This implies
$\frac{|PK|}{|PT|} = \frac{|PT|}{|PL|}, \quad \text{i.e.} \quad |PK| \cdot |PL| = |PT|^2.$
Since the power of the point $P$ with respect to the circle $k$ is $|PK| \cdot |PL| = |PA|^2$ , we can conclude that $|PA| = |PT|$ , i.e. the point $P$ is the midpoint of $\overline{AT}$ .

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