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KURSCHAK 1971 Problem 1

    A straight line cuts the side ABAB of the triangle ABCABC at C1C_1 , the side ACAC at B1B_1 and the line BCBC at A1A_1 . C2C_2 is the reflection of C1C_1 in the midpoint of ABAB , and B2B_2 is the reflection of B1B_1 in the midpoint of ACAC . The lines B2C2B_2C_2 and BCBC intersect at A2A_2 . Prove that senB1A1CsenC2A2B=B2C2B1C1\frac{sen B_1A_1C}{sen C_2A_2B} = \frac{B_2C_2}{B_1C_1} Image

    Dataset text AI-MO AoPS. Quoted with source.

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