Find the measure of angle $AMN=\theta$ Let $ABC$ be an isosceles triangle where $AB = AC$, so that $\angle BAC = 20°$. Also let $M$ be the projection of the point $C$ on the side $AB$ and $N$ a point on the side $AC$, so that $2CN = BC$. The measure of angle $AMN$ is equal to?(A:$60^o$)
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I made the drawing and marked all the angles I could find but still need to find the path...maybe an additional segment


Here is a solution using trigonometry. We will prove $IN = IM$.
First, in the triangle $INC$, we have
$$\frac{IN}{ \sin(70°)} = \frac{NC}{\sin(60)} \Longleftrightarrow IN = \frac{ \sin(70°)}{\sin(60°)}NC  \tag{1}$$
In the triangle $BMC$ and $IHC$, we have:
$$MC = \cos(10°) BC\tag{2}$$
and
$$\frac{IC}{ \sin(50°)} = \frac{HC}{\sin(120°)} \Longleftrightarrow IC = \frac{ \sin(50°)}{\sin(60°)}HC  \tag{3}$$
From $(2),(3)$, we deduce that
$$\begin{align}
IM &= MC - IC\\&=\left(2\cos(10°)-\frac{ \sin(50°)}{\sin(60°)}\right)HC\\
&=\frac{2\sin(60°)\cos(10°)-\sin(50°)}{\sin(60°)}HC\\
&=\frac{(\sin(70°)+\sin(50°))-\sin(50°)}{\sin(60°)}HC   \quad \text{because: } 2\sin(a)\cos(b)=\sin(a+b)+\sin(a-b)\\
&=\frac{ \sin(70°)}{\sin(60°)}HC  \tag{4}
\end{align}$$
From $(4),(1)$, we deduce that $IN = IM$. So the triangle $IMN$ is isosceles, then
$$90°-\theta = \theta -30° \Longrightarrow \theta = 60°$$
