In the triangle ABC, the heights AA1 and BB1 are drawn. On the side AB, points M and K are chosen so that B1K∥BC and A1M∥AC. Prove that the angle AA1K is equal to the angle BB1M.
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Tags: geometry, parallel, equal angles, angles, Kharkiv
13.03.2020 12:48
tan∠MB1B=AB1⋅BA1BChB⋅CA1BC=AB1⋅BA1hB⋅CA1 and tan∠KA1A=BA1⋅AB1AChA⋅B1CAC=BA1⋅AB1hA⋅B1C so to prove∠AA1K=∠BB1M⇔ to prove hBhA=B1CA1C It comes from a pair of familiar similar triangle △AA1C∼△BB1C
13.03.2020 13:38
The height AA1 + B1K//BC =>the angle A1EB1=90 . For the same reason,if we make D and E the intersection of A1M and BB1,AA1 and B1K,the angle A1DB1=90. So the points A1,D,E,B1 are on the same circle => the angle DA1E=the angle DB1E. B1FA1C is a parallelogram, and the angle A1AC = the angle CBB1.So: MF/FK=AC/BC=(A1C/sin( A1AC ))/(B1C/sin( CBB1 ))=(B1F/sin( A1AC ))/(A1F/sin( CBB1 ))=B1F/A1F. So MF*FA1=KF*FB1 => the points M,K,A1,B1 are on the same circle => the angle MA1K=the angle MB1K. So the angle MA1K + the angle MA1E =the angle MB1K + the angle KB1D => the angle AA1K= the angle BB1M.(there is another situation:There is no intersection between line MA1 and line KB1,this time we need turn "+" into "-") I don't konw how to send something like #2...
28.03.2024 18:14
Since B1K∥BC, hence ∠A1BB1=∠CBB1=∠BB1K. Since A1M∥AC, hence ∠B1AA1=∠CAA1=∠AA1M. We know that AB1A1B siklik, so we have ∠B1AA1=∠A1BB1, so now we have ∠AA1M=∠BB1K. Observe that ∠BMA1=180∘−∠MBA1−∠MA1B=180∘−∠CB1A1−∠B1CA1=∠CA1B1=∠A1B1KSo we have A1B1KM cyclic. Hence ∠MA1K=∠MB1K. So its proved that ∠AA1K=∠BB1M.