Problem about angle in isosceles triangle [closed]
$ABC$ 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.
Solution 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 $(\angle MDB) = 140^\circ$
$\angle BMD = 20^\circ$
$\angle BMN = 40^\circ$