angletriangles
the question
Consider the triangle $ABC$ and a point $M$ inside the triangle such that $\angle MAB = 10 ,\angle MAC = 40 ,\angle MCA = 30 $ and $\angle MBA = 20 $ . Show that triangle $ABC$ is isosceles.
my idea
the drawing
As you can see I calculated the other angles we have in the triangle…I'm preatty sure we need an auxiliar construction, but I can't figure out which one to do. I dont know how to start. Hope one of you can help me! Thank you!
This is a synthetic proof using angle chasing mostly. We need to enhance OP’s sketch by introducing point $G$, which is the reflection of the point $A$ about the segment $BC$. We join $G$ to the five points $B$, $C$, $D$, $E$ and $F$ as shown in the diagram. For brevity, let $AB=c$ and $AC=b$, $\measuredangle ABC=\beta$, and $\measuredangle BCA=\omega$. Note that, by construction, $BD=AB=c$ and $CE=AC=b$.
Since $G$ is the reflection of $A$ about $BC$, we have, $$BG=BA=c\qquad\text{and}\qquad CG=CA=b. \tag{1}$$
Furthermore, $$\measuredangle CBG=\measuredangle ABC=\beta, \quad\text{and}\tag{2a}$$ $$\measuredangle GCB=\measuredangle BCA=\omega.\qquad\space \tag{2b}$$
It can be shown that $$\measuredangle BGC =\measuredangle CAB=\pi-\measuredangle ABC-\measuredangle BCA = \pi-\beta-\omega. \tag{3}$$
It is given that $\measuredangle ABD=\dfrac{\pi}{2}$. Therefore, using (2a),we shall write, that $$\measuredangle GBD=\measuredangle ABD-\measuredangle ABC-\measuredangle CAG=\dfrac{\pi}{2}-2\beta. \tag{4a}$$
In similar vein, using (2b), we can obtain, $$\measuredangle ECG =\dfrac{\pi}{2}-2\omega. \tag{4b}$$
According to (1), both triangles $GBD$ and $ECG$ are isosceles. Since we know the apex angles of both of them (see (4a} and (4b)), we can calculate their individual base angles, i.e., $\angle DGB$ and $\angle CGE$, as shown below. $$\measuredangle DGB=\dfrac{\pi-\measuredangle GBD }{2}=\dfrac{\pi}{4}+\beta\tag{5a}$$
Similarly, we can deduce, $$\measuredangle CGE=\dfrac{\pi}{4}+\omega.\tag{5b}$$
Using (3), (5a), and (5b), we can write, $$\measuredangle EGD=\pi - \measuredangle BGC - \measuredangle DGB - \measuredangle CGE =\dfrac{\pi}{2},$$ which means $\triangle EGD $ is a right-angled triangle and, therefore, the center of its circumcircle lies at $F$. This gives us $FG=FD=FE$. Hence, both quadrilaterals $BDFG$ and $CGFE$ are kites. Since the two diagonals of a kite perpendicularly bisect each other, $BF$ and $CF$ are the angle bisectors of $\measuredangle GBD$ and $\measuredangle ECG$ respectively. Therefore, according to (4a) and (4b), we have, $$\measuredangle CBF =\measuredangle CBG +\dfrac{1}{2}\measuredangle GBD = \dfrac{\pi}{4}, \quad\text{and}$$
$$\measuredangle FCB =\measuredangle GCB+\dfrac{1}{2}\measuredangle ECG = \dfrac{\pi}{4}.\qquad\space\space$$ Therefore, $\triangle BFC$ is not only isosceles, but also right-angled.
Proceeding from the point you stopped by, let's assume $\angle MED=\alpha.$ Then, $\angle MDE=70^{\circ}-\alpha$.
By the law of sines in $\triangle AED$, we have:
$$\frac{\sin \alpha }{\sin 30^{\circ}}=\frac{DM}{AM}\times \frac{AE}{ED}.$$
But, since the segment $CX$ is the bisector, we get that: $\frac{DM}{AM}=\frac{DC}{AC}.$
Hence,
$$\frac{\sin \alpha }{\sin 30^{\circ}}=\frac{DC}{AC} \times \frac{AE}{ED} \\= \frac{\sin 60^{\circ}}{\sin 20^{\circ}}\times \frac{\sin (70^{\circ}-\alpha)}{\sin 80^{\circ}}= \frac{\sin 60^{\circ}}{\sin 20^{\circ}}\times \frac{\cos (20^{\circ}+\alpha)}{\sin 80^{\circ}}.$$
On the other hand, $\sin 20^{\circ}\sin 40^{\circ}\sin 80^{\circ}=\frac{\sqrt 3}{8}$ (look at this ), so:
$$\frac{\sin \alpha}{\cos (20^{\circ}+\alpha)}=\frac{\sin 40^{\circ}}{\cos 60^{\circ}}=\frac{\sin 40^{\circ}}{\cos (20+40)^{\circ}} \implies \alpha=40^{\circ}.$$
Best Answer
By law of sines, $$\frac{AM}{\sin 30°}=\frac{AC}{\sin 110°} \tag{1}\label{1} $$ $$\frac{AB}{\sin 150°}=\frac{AM}{\sin 20°} \tag{2}\label{2}$$ We can combine ($\ref{1}$) and ($\ref{2}$) to get $$\frac{AC\sin 30°}{\sin 110°}=\frac{AB\sin 20°}{\sin 150°}$$ Now we can apply values of $\sin 30°$, $\sin 150°$: \begin{align} AC& =2AB \left(2\sin 20°\sin 110°\right) \\ AC & =2AB(\cos(-90°)-\cos(130°)) \\ AC & =-2AB\cos(180°-50°) \\ AC & =2AB\cos 50° \tag{3} \label{3} \end{align} Again by law of cosines, $$BC^2=AC^2+AB^2-2ABAC\cos 50°$$ Now substituting from $(\ref{3})$, \begin{align} BC^2& =4(AB)^2\cos^2 50° + AB^2 - 4(AB)^2\cos^2 50° \\ BC^2 & =AB^2 \\ BC & =AB \end{align} Hence $\triangle ABC$ is isoceles.