Given a triangle triangle ABM such that |AM|=|BM| and a point C such that the oriented angle angle AMB has twice the size of angle ACB, show that |CM|=|AM|.

Madelynn Winters 2022-09-19 Answered
Given a triangle A B M such that | A M | = | B M | and a point C such that the oriented angle A M B has twice the size of A C B, show that | C M | = | A M | .
I am pretty sure that this must hold. Can somebody point me to an (elementary) proof?
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Answers (2)

Kailey Santana
Answered 2022-09-20 Author has 9 answers
Step 1
The circle around M through A (and B) intersects BC in (B and) C′. By the inscribed angle theorem, 2 A C B = A M B = 2 A C B,
Step 2
Hence A C | | A C (even though A C B = A C B + 180 o is not excluded), hence A C = A C as line and C = C as intersection of that line with BC.

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Wevybrearttexcl
Answered 2022-09-21 Author has 1 answers
Step 1
The answer is no. There are two positions for C.
If M and C are on the same side of AB, A C M = 1 2 C and C M A = π 1 2 A M B = π C and finally C A M = π A C M C M A = 1 2 C and thus A M C is isosceles and C M = A M follows.
Step 2
However, C and M could be on opposite sides of AB (so that AMBC is shaped like a kite). Then C M = A M + 2 M H where MH is the height of A M B as seen by folding AMBC along BC to get the first case.

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