OJBM

Geometry Difficulty 5.0 AIME, harder Prove it JBMO

Problem:
Let $ABC$ be a triangle inscribed in a circle $K$ . The tangent from $A$ to the circle meets the line $BC$ at point $P$ . Let $M$ be the midpoint of the line segment $AP$ and let $R$ be the intersection point of the circle $K$ with the line $BM$ . The line $PR$ meets again the circle $K$ at the point $S$ . Prove that the lines $AP$ and $CS$ are parallel.

Solution

Solution:

Figure 2
Assume that point $C$ lies on the line segment $BP$ . By the Power of Point theorem we have $MA^{2} = MR \cdot MB$ and so $MP^{2} = MR \cdot MB$ . The last equality implies that the triangles $MR$ and $MPB$ are similar. Hence $\angle MPR = \angle MBP$ and since $\angle PSC = \angle MBP$ , the claim is proved.
Slight changes are to be made if the point $B$ lies on the line segment $PC$ .

Figure 3

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.