<AA_ K=<BB_1M when B_1K//BC, A_1M//AC (Kharkiv City XI 2013 - Ukr)

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<AA_ K=<BB_1M when B_1K//BC, A_1M//AC (Kharkiv City XI 2013 - Ukr)

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#1 h Mar 13, 2020, 5:13 AM

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In the triangle $ABC$ , the heights $AA_1$ and $BB_1$ are drawn. On the side $AB$ , points $M$ and $K$ are chosen so that $B_1K\parallel BC$ and $A_1 M\parallel AC$ . Prove that the angle $AA_1K$ is equal to the angle $BB_1M$ .

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#2 h Mar 13, 2020, 9:48 AM

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$tan \angle MB_1B=\frac{AB_1 \cdot \frac{BA_1}{BC}}{h_B \cdot \frac{CA_1}{BC}}= \frac{AB_1 \cdot BA_1}{h_B \cdot CA_1}$
and $tan \angle KA_1A=\frac{BA_1 \cdot \frac{AB_1}{AC}}{h_A \cdot \frac{B_1C}{AC}} = \frac{BA_1 \cdot AB_1}{h_A \cdot B_1C}$

so to prove $\angle AA_1K=\angle BB_1M \Leftrightarrow$ to prove $\frac{h_B}{h_A}=\frac{B_1C}{A_1C}$

It comes from a pair of familiar similar triangle $\triangle AA_1C \sim \triangle BB_1C$

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modp

#4 h Mar 13, 2020, 10:38 AM • 3 Y

Y by Mango247, Mango247, Mango247

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...

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kentok

#5 h Mar 28, 2024, 3:14 PM

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Since $B_1K \parallel BC$ , hence $\angle A_1BB_1=\angle CBB_1=\angle BB_1K$ . Since $A_1M \parallel AC$ , hence $\angle B_1AA_1=\angle CAA_1=\angle AA_1M$ . We know that $AB_1A_1B$ siklik, so we have $\angle B_1AA_1=\angle A_1BB_1$ , so now we have $\angle AA_1M=\angle BB_1K$ .
Observe that
$\angle BMA_1=180^\circ-\angle MBA_1-\angle MA_1B=180^\circ-\angle CB_1A_1-\angle B_1CA_1=\angle CA_1B_1=\angle A_1B_1K$ So we have $A_1B_1KM$ cyclic. Hence $\angle MA_1K=\angle MB_1K$ . So its proved that $\angle AA_1K=\angle BB_1M$ .

This post has been edited 2 times. Last edited by kentok, Mar 28, 2024, 3:18 PM
Reason: latex

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