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
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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 $$90°-\theta = \theta -30° \Longrightarrow \theta = 60°$$
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 $\angle AMK=180^o-\angle HMB-\angle HMK=180^o-80^o-60^o=40^o$
$\therefore \angle AMN=\angle AMK+\angle KMN= 40^o+20^o=60^o$