Proving the right-angled triangle property of FDM with acute angles of $30^\circ$ and $60^\circ$

Question

I have a math question that I can't quite wrap my head around. It is:

"Let there be a triangle $ABC$. Equilateral triangles $ACE$ and $BDC$ are constructed externally on two sides. The midpoint of triangle $ACE$ is called $M$, and the midpoint of side $AF$ is called $F$. Prove that triangle $FDM$ is right-angled with acute angles of $30^\circ$ and $60^\circ$."

I have created a diagram of the figure in question, and it looks like this:

Does anybody know how I could start to solve this question?

Thanks in advance!

Answer 1

Here is another approach to the problem. I will outline the steps involved.

Extend $MF$ to $G$ s.t. $MF=FG$.

Join the lines as shown.

Prove that

(a) $BG=MC$, $\angle GBD= \angle MCD$, $BD=CD$ and hence $\Delta GBD \cong \Delta MCD.$

(b) Use the fact that $\Delta GBD \cong \Delta MCD$ to prove that $\Delta DMG$ is equilateral.

(c) Result follows because $\Delta FDM$ is just half of $\Delta DMG.$

Answer 2

Hint: You can use this property that if you construct three equilateral triangle on sides of a triangle , then the centers of construted triangle make an equilateral triangle, as can be seen in figure. The key points are:

(1):$\overset{\LARGE\frown} {O_1J}=\overset{\LARGE\frown} {O_2N}$

That means if triangle $MO_1O_2$ rotates about vertex M such that $O_2$ and $O_1$ coincide on N and J respectively, then we have:

$\angle JMN=60^o$

(2): $FH||JO_2||O_1N\Rightarrow \angle MFH=\angle MJO_2=60^o$

That means triangle MFH is equilateral and we have:

$FH=MH=MF$

This is enough to conclude that triangle MFD is right angled at F and H is mid point of MD. Only in this case angle MDF which is opposite to arc $MF=60^o$ can have measure equal to $30^o$. You probably need to show that H is mid point of MD.