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?

Question

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?

Kailey Santana · Accepted Answer

Step 1The 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 2Hence   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.

Wevybrearttexcl · Accepted Answer

Step 1The 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 2However, 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.

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|.

Answered question

Answer & Explanation

New Questions in High school geometry