For reference:
In a triangle $ABC$, the interior cevian $CM$ is drawn, so that $CM=AB$; Knowing that the measure of $\angle A= 30°$ and $\angle B =100°$. Calculate the measure of $\angle MCB$.(Answer:$40^o$)
My progress:
$MC = AB\\
\triangle AMC: \frac{sin30}{MC} = \frac{1}{2MC} = \frac{sin(100+\theta)}{AC}=\frac{sin(50-\theta)}{AM}\\
\triangle ABC: \frac{sin30}{BC}=\frac{1}{2BC} =\frac{sin100}{AC}=\frac{sin50}{AB=MC}\\
\triangle MBC: \frac{sin \theta}{MB}=\frac{sin100}{MC}=\frac{sin(80-\theta)}{BC}\\
\therefore sin(80-\theta) = \frac{sin100}{2sin50}$
….?
Best Answer
Draw $AD$ parallel and equal to $MC$. Draw the line $CE$ through $D$ parallel and equal to $AB$. Complete the parallelogram $ABCE$. Drop perpendiculars from $E$ and $B$ onto $CM$ and $AD$ respectively. Mark the foot of first one $G$ and second one $H$. The line $AH$ crosses $CM$ at $J$, and $EG$ crosses $AD$ at $K$. Quadrilateral $GKHJ$ is a rectangle. This indicates that $BH$ and $EG$ are bisectors of angles $\angle MBC$ and $\angle AEC$. Also, $BG$ bisects $\angle MBC$ and is perpendicular to $CM$, which means that triangle $BCM$ is isosceles. Therefore
$$\theta=\angle BMC=\angle BCM=\frac{180^\circ-100^\circ}2=40^\circ.$$
Update: the figure is correct. Consider this fact:
The bisectors of the angles in a parallelogram construct a rectangle.