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Prove that $\frac{AC}{DD_2}=\frac{AB}{DD_1}+\frac{BC}{DD_3}$.

Source: Moldova TST 1994

August 8, 2023

geometry

Problem Statement

Inside the triangle

DD_1D_3

the cevian

DD_2

is constructed. Perpendiculars from

D_1, D_2

and

D_3

to lines

DD_1, DD_2

and

DD_3

, respectively, intersect in points

A,B

and

C

such that

AB\perp DD_1, AC\perp DD_2, BC\perp DD_3

. Prove that

\frac{AC}{DD_2}=\frac{AB}{DD_1}+\frac{BC}{DD_3}