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

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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!

Best Answer

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.

Best Answer

Related Solutions

In an acute angled triangle $ABC$, the angle bisector $AL$, altitude $BH$ and perpendicular bisector of $AB$ are concurrent. the $\angle BAC$

In triangle $ABC$, $\angle A=50^{\circ}$. Point $D$ is constructed inside the triangle such that $\angle BDC=130^{\circ}$.

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