Prove that △ MKP is regular.

Question

We have a △ ABC with angle ACB equal to 60° and an orthocenter H. The circle with diameter CH intersects with AC in point M and with BC in point K. P is the midpoint of AB. Prove that △ MKP is regular.

You can find a diagram here

Answer 1

Let $c$ be the circle with diameter $CH$. Then, since $CH$ is a diameter of $c$, angles $\angle \, CMH = \angle \, CKH = 90^{\circ}$ and so lines $HM$ and $HK$ are orthogonal to $AC$ and $BC$ respectively. On the other hand, $H$ is the orthocenter of triangle $ABC$, so the lines $BH$ and $AH$ are orthogonal to $AC$ and $BC$ respectively. But there is only one line through $H$ orthogonal to $AC$ and by construction that line is $HK$. Therefore, points $B, H, M$ are collinear. Analgoously, points $A, H, K$ are collinear and so $BM$ and $AK$ are two of the altitudes of triangle $ABC$. Then $\angle \, AMB = \angle \, AKB = 90^{\circ}$ which means that quadrilateral $ABKM$ is inscribed in a circle, call it $c_1$, and $AB$ is its diameter. Therefore, $P$ is the center of $c_1$. Consequently, $PM = PK = PA = PB = \frac{1}{2} AB$ are radii of $c_1$, which means triangle $MKP$ is isosceles. Let $\angle \, ACH = \angle MCH = \alpha$. Then, since quadrilateral $MHKC$ is inscribed in circle $c$, we get $\angle \, MKH = \angle MCH = \alpha$. On the other hand, since quadrilateral $ABKM$ is inscribed in circle $c_1$, we get $\alpha = \angle \, MKH = \angle \, MKA = \angle \, MBA = \angle \, MBP$. Since $P$ is a center of $c_1$, triangle $BMP$ is isosceles and $\angle \, PMH = \angle \, PMB = \angle \, MBP = \alpha$. On the other hand, as $MHKC$ is inscribed in $c$, we get $\angle \, HMK = \angle \, HCK = 60^{\circ} - \alpha$. Therefore, $\angle \, PMK = \angle \, PMH + \angle \, HMK = \alpha + 60^{\circ} - \alpha = 60^{\circ}.$ Consequently, triangle $PMK$ is isosceles with one angle of $60^{\circ}$ and is therefore equilateral.

Answer 2

By angle chasing, it is easy to prove that $M$ and $K$ are just the feet of the latitudes from $A$ and $B$.
Let we embed the construction in the complex plane by setting $C=0$, $A=1$ and $B=\lambda\omega$ with $\omega=e^{\pi i/3}$. We have $$ M = \frac{\lambda}{2},\qquad K=\frac{\omega}{2},\qquad P=\frac{1+\lambda \omega}{2}$$ and it is straightforward to check that $\|M-K\|^2 = \|M-P\|^2 = \|P-K\|^2$, so $PMK$ is an equilateral triangle.