MathDB
sen B_1A_1C / sen C_2A_2B =B_2C_2 / B_1C_1

Source: 1971 Hungary - Kürschák Competition p1

October 15, 2022
geometrytrigonometryratio

Problem Statement

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} https://cdn.artofproblemsolving.com/attachments/3/8/774da81495df0a0f7f2f660ae9f516cf70df06.png