Finding angle in triangle

Question

I have a triangle $\triangle ABC$ with vertices $A$, $B$ and $C$ with $\angle BAC = 45^\circ$ and $\angle ABC = 30^\circ$. Point $M$ is the center of the $BC$. Find $\angle AMC$.

This is not my homework, it's a little girl's, so she can't use trigonometric equations or anything after special lines in triangles.

Thanks in advance!

Answer 1

Let $X$ on $AB$ be the foot of the altitude through $C$.

• Since $\angle BAC = 45^\circ$, $\angle ACX = 45^\circ$, $\triangle AXC$ is isoceles and $AX = CX$.

• Since $\angle CBA = 30^\circ$, $\angle XCB = 60^\circ$, $\triangle XBC$ is half of an equilateral triangle and $CX : BC = 1 : 2$.

• Since $M$ is midpoint of $BC$, $CX = CM$ and $\triangle XCM$ is equilateral.

• This implies $XM = CM = BM$, $\triangle XBM$ is isosceles and $\angle BXM = \angle MBX = 30^\circ$.

• Since $AX = CX = XM$, $\triangle AXM$ is also isosceles with $\angle XAM = \angle AMX = \frac12 \angle BXM = 15^\circ$.

From this, we find

$$\begin{align} \angle CMA &= 180^\circ - \angle AMX - \angle XMB \\ &= 180^\circ - \angle AMX - (180^\circ -\angle BXM - \angle MBX )\\ &= \angle BXM + \angle MBX - \angle AMX\\ &= 30^\circ + 30^\circ - 15^\circ\\ &= 45^\circ\end{align}$$

Answer 2

Let $\measuredangle AMC=x$.

Hence, $$\frac{MC}{\sin(75^{\circ}-x)}=\frac{AC}{\sin{x}}$$ and $$\frac{BM}{\sin(x-30^{\circ})}=\frac{AB}{\sin{x}},$$ which gives $$\frac{\sin(x-30^{\circ})}{\sin(75^{\circ}-x)}=\frac{\sin30^{\circ}}{\sin105^{\circ}}$$ or $$\sin{x}\cos30^{\circ}-\frac{1}{2}\cos{x}=2\sin15^{\circ}(\sin75^{\circ}\cos{x}-\cos75^{\circ}\sin{x})$$ or $$\tan{x}=\frac{2\sin15^{\circ}\sin75^{\circ}+\frac{1}{2}}{\cos30^{\circ}+2\sin15^{\circ}\cos75^{\circ}}$$ or $$tan{x}=\frac{\cos60^{\circ}-\cos90^{\circ}+\frac{1}{2}}{\cos30^{\circ}+\sin90^{\circ}-\sin60^{\circ}}$$ or $$\tan{x}=1,$$ which gives $$x=45^{\circ}.$$

Answer 3

Let $A_1$ be on circumcircle of $\Delta ABC$, such that $A_1$ placed on an arc $BAC$ and $\measuredangle MA_1C=30^{\circ}$.

Let $BK$ be an altitude of $\Delta A_1BC$ and $MK$ is an altitude of $\Delta MA_1C$.

Since $M$ is a midpoint of $BC$, we obtain $MN=\frac{1}{2}BK$ and since $\measuredangle MA_1C=30^{\circ}$,

we see that $MN=\frac{1}{2}A_1M$.

Thus, $A_1M=BK$.

But $\measuredangle BA_1C=\measuredangle BAC=45^{\circ}$,

which gives $BK=A_1K$ and from here $A_1M=A_1K$,

which says that $\measuredangle MKA_1=\frac{180^{\circ}-30^{\circ}}{2}=75^{\circ}$.

But $MK=MB=MC$, which says that $\measuredangle BCK=75^{\circ}$, which gives $\measuredangle BCA_1=105^{\circ}$.

But we know that $\measuredangle BCA=105^{\circ}$, which gives $A\equiv A_1$.

Thus, $\measuredangle BAM=15^{\circ}$ and $$\measuredangle AMC=\measuredangle ABM+\measuredangle BAM=30^{\circ}+15^{\circ}=45^{\circ}$$ and we are done!