Proof regarding altitudes of a triangle and a midpoint of one of its sides

Question

Let $\triangle ABC$ be a triangle with altitudes $\overline{AH}$ and $\overline{BK}$. Consider the axis of the segment $ \overline{HK}$. Let $M$ be the point of intersection between the axis and the side $\overline{AB}$.

I have to prove that $\overline{AM} \cong \overline{MB}$.

If the triangle $\triangle ABC$ is isosceles the the proof is easy; but in the general case I don't know how to proceed. I have noticed that $\overline{MH} \cong \overline{MK}$ but I can't conclude.

Thank you for your help!

Answer 1

Notice that both triangles $\Delta ABH$ and $\Delta BKA$ have a right angle $(=90°)$ and share the basis $\overline {AB}$.

By the inversion of Thales's Theorem, they are inscribed in the same circle $\omega$ with diameter $\overline {AB}$.

$\overline {KH}$ is, therefore, one chord of $\omega$, implying that the perpendicular bisector of $\overline {KH}$ passes through the circle's center, i.e. $\overline {AB}$'s midpoint $M$, see why?

Since $\overline {AB}$ is the diameter, the midpoint $M_{AB}$ of $\overline {AB}$ is $\omega$'s centre. Notice now that points $H$ and $K$ both lie on the circumference. Since $$\overline {M_{AB}K}=\overline {M_{AB}H}=\omega \text{'s radius}$$ which implies that $M_{AB}$ lies on the perpendicular bisector of $\overline {HK}$

It follows that

$$\overline {AM}\cong \overline {MB}$$

Answer 2

Since $$\angle AKB = \angle AHB = 90$$ we see that $A,B,H,K$ are concyclic. The circle around $A,B,H,K$ has diameter $AB$ (by Thales theorem) and thus a conclusion.