Subfield: Geometry


Question: In the diagram, $C$ lies on $B D$. Also, $\triangle A B C$ and $\triangle E C D$ are equilateral triangles. If $M$ is the midpoint of $B E$ and $N$ is the midpoint of $A D$, prove that $\triangle M N C$ is equilateral.

<img_3743>

Solution:
Consider $\triangle B C E$ and $\triangle A C D$.

<img_3765>

Since $\triangle A B C$ is equilateral, then $B C=A C$.

Since $\triangle E C D$ is equilateral, then $C E=C D$.

Since $B C D$ is a straight line and $\angle E C D=60^{\circ}$, then $\angle B C E=180^{\circ}-\angle E C D=120^{\circ}$.

Since $B C D$ is a straight line and $\angle B C A=60^{\circ}$, then $\angle A C D=180^{\circ}-\angle B C A=120^{\circ}$.

Therefore, $\triangle B C E$ is congruent to $\triangle A C D$ ("side-angle-side").

Since $\triangle B C E$ and $\triangle A C D$ are congruent and $C M$ and $C N$ are line segments drawn from the corresponding vertex ( $C$ in both triangles) to the midpoint of the opposite side, then $C M=C N$.

Since $\angle E C D=60^{\circ}$, then $\triangle A C D$ can be obtained by rotating $\triangle B C E$ through an angle of $60^{\circ}$ clockwise about $C$.

This means that after this $60^{\circ}$ rotation, $C M$ coincides with $C N$.

In other words, $\angle M C N=60^{\circ}$.

But since $C M=C N$ and $\angle M C N=60^{\circ}$, then

$$
\angle C M N=\angle C N M=\frac{1}{2}\left(180^{\circ}-\angle M C N\right)=60^{\circ}
$$

Therefore, $\triangle M N C$ is equilateral, as required.
We prove that $\triangle M N C$ is equilateral by introducing a coordinate system.

Suppose that $C$ is at the origin $(0,0)$ with $B C D$ along the $x$-axis, with $B$ having coordinates $(-4 b, 0)$ and $D$ having coordinates $(4 d, 0)$ for some real numbers $b, d>0$.

Drop a perpendicular from $E$ to $P$ on $C D$.

<img_3660>

Since $\triangle E C D$ is equilateral, then $P$ is the midpoint of $C D$.

Since $C$ has coordinates $(0,0)$ and $D$ has coordinates $(4 d, 0)$, then the coordinates of $P$ are $(2 d, 0)$.

Since $\triangle E C D$ is equilateral, then $\angle E C D=60^{\circ}$ and so $\triangle E P C$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle and so $E P=\sqrt{3} C P=2 \sqrt{3} d$.

Therefore, the coordinates of $E$ are $(2 d, 2 \sqrt{3} d)$.

In a similar way, we can show that the coordinates of $A$ are $(-2 b, 2 \sqrt{3} b)$.

Now $M$ is the midpoint of $B(-4 b, 0)$ and $E(2 d, 2 \sqrt{3} d)$, so the coordinates of $M$ are $\left(\frac{1}{2}(-4 b+2 d), \frac{1}{2}(0+2 \sqrt{3} d)\right)$ or $(-2 b+d, \sqrt{3} d)$.

Also, $N$ is the midpoint of $A(-2 b, 2 \sqrt{3} b)$ and $D(4 d, 0)$, so the coordinates of $N$ are $\left(\frac{1}{2}(-2 b+4 d), \frac{1}{2}(2 \sqrt{3} b+0)\right)$ or $(-b+2 d, \sqrt{3} b)$.

To show that $\triangle M N C$ is equilateral, we show that $C M=C N=M N$ or equivalently that $C M^{2}=C N^{2}=M N^{2}$ :

$$
\begin{aligned}
C M^{2} & =(-2 b+d-0)^{2}+(\sqrt{3} d-0)^{2} \\
& =(-2 b+d)^{2}+(\sqrt{3} d)^{2} \\
& =4 b^{2}-4 b d+d^{2}+3 d^{2} \\
& =4 b^{2}-4 b d+4 d^{2} \\
C N^{2} & =(-b+2 d-0)^{2}+(\sqrt{3} b-0)^{2} \\
& =(-b+2 d)^{2}+(\sqrt{3} b)^{2} \\
& =b^{2}-4 b d+4 d^{2}+3 b^{2} \\
& =4 b^{2}-4 b d+4 d^{2} \\
M N^{2} & =((-2 b+d)-(-b+2 d))^{2}+(\sqrt{3} d-\sqrt{3} b)^{2} \\
& =(-b-d)^{2}+3(d-b)^{2} \\
& =b^{2}+2 b d+d^{2}+3 d^{2}-6 b d+3 b^{2} \\
& =4 b^{2}-4 b d+4 d^{2}
\end{aligned}
$$

Therefore, $C M^{2}=C N^{2}=M N^{2}$ and so $\triangle M N C$ is equilateral, as required.



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