MathDB

2006.8.3

Part of Kyiv City MO Juniors 2003+ geometry

Problems(1)

KL=LB, perpendiculars in right triangle (Kyiv City Olympiad 2006 8.3)

Source:

6/28/2020
On the legs AC,BCAC, BC of a right triangle △ABC\vartriangle ABC select points MM and NN, respectively, so that ∠MBC=∠NAC\angle MBC = \angle NAC. The perpendiculars from points MM and CC on the line ANAN intersect ABAB at points KK and LL, respectively. Prove that KL=LBKL=LB.
(O. Clurman)
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