Problem about angle in isosceles triangle

Question

$ABC$ is an isosceles triangle, $AB=AC$, and $\measuredangle A=20^{\circ}$. For point $M$ on $AB$ and $N$ on $AC$, we have $AM=NC=BC$. Compute the $\measuredangle BMN$.

In a similar problem (with only one point on one side) the key step was to construct an equilateral triangle on the other side. I tried using that in this problem too, but I didn't manage to solve it.

Answer 1

Locate $D$ such that $\triangle BDC$ forms equilateral triangle.

$\angle ABC = 80^\circ$

$\angle MBD = 20^\circ$

$BD = AM, \angle MBD = \angle NAM, MB = AN$

$\triangle BDM \cong \triangle AMD$

MD = MN

$\triangle MDN$ is an isosceles triangle.

$\angle DMN = 180 - \angle DMB- \angle AMN = \angle MBD = 20$ degrees

$\triangle DCN$ is an isosceles triangle with $20$ degree vertex.

The known angles around $D$ sum to $80^\circ+80^\circ+60^\circ=220^\circ$

The reamianing angle $(\angle MDB) = 140^\circ$

$\angle BMD = 20^\circ$

$\angle BMN = 40^\circ$

Answer 2

you can use a $4\times4$ matrix to solve it, where

$$ let ~a = \measuredangle AMN, b = \measuredangle ANM, c = \measuredangle BMN, d = \measuredangle CNM $$

you have

$$ a+c = 180^{\circ} \\ b + d = 180^{\circ} \\ a + b = 160^{\circ} \\ c + d = 200^{\circ} $$