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

 A: 
Let $CN=x$, then $BH=HC =x$.
In the figure, $K$ is constructed so that $KN//BC$ and $KH//CN$
$\therefore CNKH$ is a parallelogram
$\because CN=CH=x$,
$\therefore NK=HK=x$
Note that $\angle CMB=90^o$ and $H$ is the mid-point of $BC$ implies that $ HM=HB=HC=x$.
Thus $HM=HK$ ------ (1)
$\because \angle KHC=100^o$ and $\angle MHB=20^o$
$ \therefore \angle KHM=60^o$ ----- (2)
(1), (2) implies that $\Delta MHK$ is an equilateral triangle.
Hence $MK=HK$-----(3)
$\because HK=KN$ ------(4)
(3), (4) $\implies MK=KN$
Since $\angle HKN=\angle NCH=80^o$, $\angle MKN=60^o+80^o=140^o$
Thus$\angle KMN=\frac{180^o-140^o}{2}=20^o$
Also $\angle AMK=180^o-\angle HMB-\angle HMK=180^o-80^o-60^o=40^o$
$\therefore \angle AMN=\angle AMK+\angle KMN=
40^o+20^o=60^o$
A: 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
$$90°-\theta = \theta -30° \Longrightarrow \theta = 60°$$
