angleeuclidean-geometrygeometrytriangles
For reference: In a triangle ABC a $m \measuredangle A = 2m \measuredangle C$.
Plot the median BM such that $\measuredangle AMB=45º$. Calculate the $m \measuredangle C$.
My progress:
$Trace DM \perp AC\\
EXtend~AB~ to~D\\
\text{mark point E (circuncentro) on the straight DM in the } \triangle ADC\\\overset{\LARGE{\frown}}{CD}=2x \therefore \measuredangle AEC = 4x $
it remains to demonstrate that EAM =2x???
$\triangle ABC$ is isosceles. Therefore,
$\measuredangle BAC = \measuredangle C \implies 2\alpha+\alpha =120^\circ\implies \alpha =40^\circ$
Therefore $\measuredangle ABC = 180^\circ - 160^\circ = 20^\circ$
Draw $AX$ such that $\measuredangle XAK = 30^\circ$.
Draw $TX$.
$\measuredangle CDB = 120^\circ ~~\therefore \measuredangle DIA = 90^\circ$
Since $CI$ is angle bisector $IA = IX$
$\triangle ITA \cong \triangle IXT ~(L.A.L) \rightarrow \measuredangle AXT = 60^\circ$
Therefore $\triangle ATX$ is equilateral.
$\measuredangle KXB = 180 ^\circ -110^\circ= 70^\circ$
$AK$ is angle bisector and the height of triangle $ATX$
$ \therefore AKT = 90^\circ$
$TK = KX \rightarrow \triangle KTB \cong \triangle KXB~ (L.A.L)$
Therefore $\triangle TBX$ is isosceles.
$\measuredangle B = 180^\circ -140^\circ =40^\circ$
$\therefore \boxed{\color{red}x=40^\circ-20^\circ=20^\circ}$
Yes, you are correct, $\measuredangle b$ is $90^\circ$, but not $\measuredangle a$.
Hint: Consider the pentagon (sum of interior angles=$540^\circ$). Can you take it from here?
Best Answer
Another way.
Let $G$ be a mid-point of $AB$ and $BE$ be an altitude of $\Delta ABC$.
Thus, since $AG=GE=GB,$ $BE=EM$ and
$GE=EM$ (because $GM||BC$, $\measuredangle GMA=\measuredangle C$ and $\measuredangle GEA=\measuredangle A=2\measuredangle C$ ), we obtain: $$EB=EM=EG=\frac{1}{2}AB,$$ which gives $$\measuredangle BAC=30^{\circ}$$ and $$\measuredangle C=15^{\circ}.$$