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6

Part of 1994 Moldova Team Selection Test

Problems(1)

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

Source: Moldova TST 1994

8/8/2023
Inside the triangle DD1D3DD_1D_3 the cevian DD2DD_2 is constructed. Perpendiculars from D1,D2D_1, D_2 and D3D_3 to lines DD1,DD2DD_1, DD_2 and DD3DD_3, respectively, intersect in points A,BA,B and CC such that AB⊥DD1,AC⊥DD2,BC⊥DD3AB\perp DD_1, AC\perp DD_2, BC\perp DD_3. Prove that ACDD2=ABDD1+BCDD3\frac{AC}{DD_2}=\frac{AB}{DD_1}+\frac{BC}{DD_3}.
geometry