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:$60^o$)
Follow my progress..
I made the drawing and marked all the angles I could find but still need to find the path…maybe an additional segment
Best Answer
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 $\angle AMK=180^o-\angle HMB-\angle HMK=180^o-80^o-60^o=40^o$
$\therefore \angle AMN=\angle AMK+\angle KMN= 40^o+20^o=60^o$