# Prove triangle made from two altitudes and midpoint is isosceles

In triangle ABC, AH and BK are altitudes. M is the midpoint of AB. Prove that triangle MHK is isosceles.

All I can see is that the angles formed where the altitudes intersect are equal, and since each altitude makes a right angle with the opposite side, angle KAH and angle HBK must be congruent.

Note that we only have congruent triangle and parallel line theorems (not even similar triangles, yet).

Triangles $ABH$ and $ABK$ are right triangles whose hypothenuse is $AB$. So they have the same circumcircle, whose diameter is $AB$. As a consequence, considering that $M$ is the midpoint of the diameter $AB$, we get that $MH=MK$ because they are radii of the same circumference, and then $MHK$ is isosceles.

Without using circles: we can show that, in any right triangle, the median drawn to the hypotenuse is equal to half the hypotenuse. To show it for the right triangle $AHB$, let us draw a line $ME$ starting from the midpoint $M$, parallel to the leg $AH$, and ending to the intersection point $E$ with the other leg $HB$. We know that the angle $AHB$ is right. The angles $MEB$ and $AHB$ are congruent, since they are corresponding angles if we consider the parallel lines $AH$ and $ME$, and the transverse line $HB$. So, the angle $MEB$ is a right angle.

Because the line $ME$ starts from the midpoint $M$ and is parallel to $AH$, it divides the leg $HB$ in two congruent segments $HE$ and $EB$ with equal length. These considerations show that the triangles $MEH$ and $MEB$ are right triangles, have congruent legs $HE$ and $EB$, and a common leg $ME$. This means that these triangles are congruent. We then get that the segments $MH$ and $MB$ are also congruent, since they are corresponding sides (hypothenuses) of these triangles. So we have shown that, in the right triangle $AHB$, the median $MH$ is equal to half the hypotenuse $AB$.

Applying the same procedure to the right triangle $BKA$, we get that $MK$ is congruent with $MA$, so that it also equals half of the hypothenuse $AB$.

We then conclude that $MH=MK$, and so the triangle $MHK$ is isosceles.

• I'm afraid that we can't use circles yet. Only congruent triangle and parallel line theorems. Also, I don't see how you get that AH=BK. They could be any secants. – Jeff Aug 26 '14 at 0:32
• I edited my answer adding an alternative proof that does not use circles, and that is based on congruences and the parallel line theorem. – Anatoly Aug 26 '14 at 2:10
• Good answer. Can you clarify a little how we know that E is the midpoint of HB? – Jeff Aug 26 '14 at 2:44
• Draw through $H$ the parallel to $AB$ and prolong $ME$ beyond $E$ until you intersect that line in $F$. We can show that triangles $EFH$ and $EMB$ are congruent using parallel line theorems. – Anatoly Aug 26 '14 at 6:38
• Did that. I can see that $\angle HEF = \angle MEB$ (vertical angles) and that $\angle HFE = \angle EMB$ (alternate interior angles). But I do not see how to equate two sides of the triangles to prove congruence. – Jeff Aug 26 '14 at 14:28