In triangle ABC, ∠ A = 20 º and A M = C N = C B . Find angle ∠ M B N ?.

Question

In triangle ABC,   ∠  A  =  20      º   and   A  M  =  C  N  =  C  B . Find angle   ∠  M  B  N ?.

Cristal Roth · Accepted Answer

Step 1We draw an equilateral triangle with side AB as shown. Then   △  P  A  C is isosceles with   ∠  P  A  C  =      40    ∘   . That leads to   ∠  B  P  C  =      10    ∘   . Also,   ∠  P  B  C  =      20    ∘   .Now   △  P  B  C  ≅  △  B  A  M (by S-A-S)So it follows that   ∠  A  B  M  =      10    ∘   .  ∴  ∠  M  B  N  =      80    ∘    −      50    ∘    −      10    ∘    =      20    ∘

Friegordigh7r7 · Accepted Answer

Step 1Using angle chasing,   ∠  A  C  B  =  ∠  A  B  C  =            180              º            −      20              º              2    =  80      º   . Since   Δ  N  C  B is isosceles,   ∠  B  N  C  =  ∠  N  B  C  =  50      º   .Now we have to make use of the information   A  M  =  C  N  =  C  B , focusing in on AM in particular. Thus, let us reflect triangle ABC horizontally around the middle, so that base AB stays in the same position. Thus   C  N  =  C  B  =  A      C    ′    =  A  M . Since   ∠      B    ′    =  80      º   by symmetry,   ∠  C  A  M  =  80      º    −  20      º    =  60      º   . Since   Δ  C  A  M is also iscoceles, this triangle must be equilateral! Therefore   ∠  A      C    ′    M  =  ∠  A  M      C    ′    =  60      º   as well.Again by symmetry, we have that   A  E  =  E  B . Thus   ∠  E  B  A  =  20      º   also and   ∠  A  E  B  =  140      º   . Now if we construct D such that   A  M  =  M  D and D lies on AB, then   ∠  M  D  A  =  20      º   as well. Thus by AA similarity,   Δ  A  M  D  ∼  Δ  A  E  B .

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