MathLabs

Problem 2

Let ABCABC be a triangle with circumcenter OO. The points PP and QQ are interior points of the sides CACA and ABAB, respectively. Let KK, LL, and MM be the midpoints of the segments BPBP, CQCQ, and PQPQ, respectively, and let Γ\Gamma be the circle passing through KK, LL, and MM. Suppose that the line PQPQ is tangent to the circle Γ\Gamma. Prove that OP=OQOP = OQ.
Step 3 of 4: Similar triangles give the product identity
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△AQP∼△MLK  ⟹  AQ⋅QB=AP⋅PC\triangle AQP \sim \triangle MLK \implies AQ\cdot QB = AP\cdot PC
Detailed analysis

Triangle AQPAQP has ∠QAP=∠BAC=∠KML\angle QAP=\angle BAC=\angle KML (Step 1) and ∠AQP=∠MLK\angle AQP=\angle MLK (Step 2), so △AQP∼△MLK\triangle AQP\sim\triangle MLK (correspondence A↔M, Q↔L, P↔KA\leftrightarrow M,\ Q\leftrightarrow L,\ P\leftrightarrow K) by AA. This similarity gives AQAP=MLMK\dfrac{AQ}{AP}=\dfrac{ML}{MK}. Combining with MLMK=PCQB\dfrac{ML}{MK}=\dfrac{PC}{QB} from Step 1, we get AQAP=PCQB\dfrac{AQ}{AP}=\dfrac{PC}{QB}, i.e. AQ⋅QB=AP⋅PCAQ\cdot QB=AP\cdot PC.