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ratios of segments and ratios of cosines determine concurrence of some lines

Source: Romanian District Olympiad 2017, Grade IX, Problem 1

October 10, 2018
geometrytrigonometryratio

Problem Statement

Let A1,B1,C1 A_1,B_1,C_1 be the feet of the heights of an acute triangle ABC. ABC. On the segments B1C1,C1A1,A1B1, B_1C_1,C_1A_1,A_1B_1, take the points X,Y, X,Y, respectively, Z, Z, such that {C1XXB1=bcos⁡∠BCAccos⁡∠ABCA1YYC1=ccos⁡∠BACacos⁡∠BCAB1ZZA1=acos⁡∠ABCbcos⁡∠BAC. \left\{\begin{matrix}\frac{C_1X}{XB_1} =\frac{b\cos\angle BCA}{c\cos\angle ABC} \\ \frac{A_1Y}{YC_1} =\frac{c\cos\angle BAC}{a\cos\angle BCA} \\ \frac{B_1Z}{ZA_1} =\frac{a\cos\angle ABC}{b\cos\angle BAC} \end{matrix}\right. . Show that AX,BY,CZ, AX,BY,CZ, are concurrent.
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