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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).

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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.

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  • $\begingroup$ 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. $\endgroup$ – Jeff Aug 26 '14 at 0:32
  • $\begingroup$ I edited my answer adding an alternative proof that does not use circles, and that is based on congruences and the parallel line theorem. $\endgroup$ – Anatoly Aug 26 '14 at 2:10
  • $\begingroup$ Good answer. Can you clarify a little how we know that E is the midpoint of HB? $\endgroup$ – Jeff Aug 26 '14 at 2:44
  • $\begingroup$ 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. $\endgroup$ – Anatoly Aug 26 '14 at 6:38
  • $\begingroup$ 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. $\endgroup$ – Jeff Aug 26 '14 at 14:28

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