Elementary problem of geometry involving equilateral triangles

Question

I have the following exercise of elementary geometry: Given a triangle $ABC$, and draw (to the outside of this triangle) two equilateral triangles $ABE$ and $ACF$. Let $M,P$ be the midpoints of $BC,EF$ respectively and $H$ the projection of $A$ on $EF$. Prove that $MP=MH$.

Sorry if this problem bothers you, but I do not have any idea to do it, except using coordinates. Does someone have any idea?

Answer 1

Denote by $L$ and $N$ the midpoints of segments $BE$ and $CF$ respectively. Then since triangles $BAE$ and $CAF$ are equilateral, $$\angle \, ALE = \angle \, AHE = 90^{\circ}$$ $$\angle \, ANF = \angle \, AHF = 90^{\circ}$$ so quads $LAHE$ and $NAHF$ are inscribed in circles respectively, so $$\angle \, LHA = \angle \, LEA = 60^{\circ}$$ $$\angle \, NHA = \angle \, NFA = 60^{\circ}$$ and so $$\angle \, LHN = \angle \, LHA + \angle \, NHA = 120^{\circ}$$

Perform a $60^{\circ}$ counterclockwise rotation around point $A$. Then, since triangles $BAE$ and $CAF$ are equilateral, point $E$ is mapped to point $B$ and point $C$ is mapped to point $F$. Thus, triangle $EAC$ is mapped to triangle $BAF$ so they are congruent and moreover segment $EC$ is mapped to segment $BF$, which means that $EC = BF$ and the two of them intersect at angle $60^{\circ}$ (as well as $120^{\circ}$ as an external angle).

The segments $MN$ and $LP$ are mid-segments in triangles $BCF$ and $BEF$ respectively and so $MN$ and $LP$ are both parallel to $BF$ and $MN = \frac{1}{2} \, BF = LP$. Analogously, segments $PN$ and $LM$ are both parallel to $EC$ and $PN = \frac{1}{2} \, EC = LM$. But as already shown, $EC = BF$ and they intersect at angle $60^{\circ}$, the quadrilateral $LMNP$ is a rhombus with angles $$\angle\, LPN = \angle \, LMN = 120^{\circ}$$ $$\angle\, MNP = \angle \, MLP = 60^{\circ}$$ This means that triangles $MNP$ and $MLP$ are equilateral so $$MP = MN = ML$$

Furthermore, observe that $\angle \, LHN = \angle\, LPN = 120^{\circ}$ so the quad $LHPN$ is inscribed in a circle $k,$ which means that the point $H$ lies on the circle $k$ circumscribed around triangle $LPN$. However, the circle with center $N$ and radius $MP = MN = ML$ passes through the three points $L, P, N$ and so it is the circle $k$ circumscribed around triangle $LPN$ and so $H$ lies on $k$ and since $M$ is the center of circle $k$, $$MH = MP = MN = ML$$

Answer 2

My work is quite similar to that of @futurologist .

Fact-1 If L is the midpoint of BE, then $\angle BLA = 90^0$. This means the red dotted circle can be formed. As a consequence, $\angle LHA = \angle BEA = 60^0$. Similarly, $\angle NHA = \angle CFA = 60^0$.

Fact-2 After drawing an extra equilateral triangle BCQ outside of the line BC, we obtained the Fermat Point, X such that the blue lines AQ, BF, and CE are equal in length. In addition, they cut each other at angles of $60^0$ (i. e. $\angle FXC = 60^0$).

Fact-3 LMNP is a rhombus (because each side is equal to half of a blue line according to the midpoint theorem). In addition, $\angle PNM$, $\angle MPN$, $\angle LPM$, and $\angle PLM$ are all equal to $\angle FXC = 60^0$ (because UXVN is also a parallelogram). This means M is the center of the circle passing through L, P, and N.

Putting the result of Fact-1 in, we can say that L, H, P and N are con-cyclic. Required result follows.

Answer 3

Let us put everything in complex plane. Clearly $B,C\in \mathbb{R}$ and $A\in i\mathbb{R}$. Let $O$ be origin.

Instead of proving $MH=MP$ we will prove $OP=PM$ or $Re(P) ={1\over 2}M$.

Let $\varepsilon = \cos {\pi\over 3}+ i \sin {\pi\over 3}$. Then $-\varepsilon ^{-1}= \varepsilon ^2 = \varepsilon -1$ and

\begin{eqnarray} F &=& A+\varepsilon (C-A)\\ E &=& A-\varepsilon^2 (B-A) \end{eqnarray} Since $M={1\over 2}(B+C)$ and

$$P ={1\over 2}(E+F) = {1\over 2}\Big(A(2-\varepsilon +\varepsilon ^2) +\varepsilon C - \varepsilon ^2 B \Big) = {1\over 2}\Big(A +\varepsilon C - \varepsilon ^2 B \Big)$$ we have \begin{eqnarray} Re(P) &=& {1\over 2}(P+\overline{P}) \\ &=& {1\over 4}\Big(A +\varepsilon C - \varepsilon ^2 B + \overline{A +\varepsilon C - \varepsilon ^2B} \Big)\\ &=& {1\over 4}\Big(A +\varepsilon C - \varepsilon ^2 B -A +\overline{\varepsilon }C - \overline{\varepsilon }^2B \Big)\\ &=& {1\over 4}\Big((\underbrace{\varepsilon +\overline{\varepsilon }}_{=1})C - (\underbrace{\varepsilon ^2 + \overline{\varepsilon }^2}_{-1})B \Big)\\ &=& {1\over 4}(B+C)\\ &=& {1\over 2} M \end{eqnarray}