Greatest angle chasing ever

Geometry Level 5

If $\Delta ABC$ is isosceles with $AB = AC$ . Let $M$ be a point interior of the triangle and $\angle MBC = { 25 }^{ \circ }$ , $\angle MCB\ = { 30 }^{ \circ }$ , $\angle MBA = { 10 }^{\circ }$ , $\angle MCA = { 5 }^{ \circ }$ . Find angle $\angle AMC$ .

The answer is 150.

6 solutions

Pulkit Deshmukh
Sep 25, 2015

Draw angle bisector AF and reflect M upon AF to M'

From symmetry, $\angle MAM'=\angle ABM'=\angle ACM= 5^{\circ}$ as MM' is parallel to BC we get $\Delta AM'B= \Delta MM'B$ hence $\angle AM'B=\angle AMC=150^{\circ}$