# Prove triangles formed by two midpoints and an altitude are congruent

Triangle ABC has altitude BH. M is the midpoint of AB, and N is the midpoint of CB. Prove triangle MBN is congruent to triangle MHN.

Can we say that MN bisects BH? If so, why?

If MN bisects BH (at point X), do they form right angles? If so, why? Is a segment between two midpoints parallel to the triangle's opposite side?

If we have both of those, then triangle BXN is congruent to triangle HXN. Then angle HBN = angle BHN and BN = HN. Also, triangles BXM = HXM, so angles BHM = HBM and sides BM = HM. Then we can prove using Side-angle-side.

Note: Only theorems about triangle congruence and parallel lines are available (no similar triangles).

• "Can we say that MN bisects BH? If so, why?" Yes. Because BMN and BAC are similar (two proportional sides and a common interior angle). So Triangle BXM and BHA are similar (two angles are equal) so $BX$ is proportional to $BH$ as $BM$ is proportional to $BA$. Sep 7, 2016 at 0:25
• "Is a segment between two midpoints parallel to the triangle's opposite side?" Yes. Because they form similar triangles so the form equal interior angles. "If MN bisects BH (at point X), do they form right angles? If so, why? " Yes, because the lines are parallel. Sep 7, 2016 at 0:28
• Oh. No similar triangles. .... Okay. Let X be midpoint BH. Construct perp biscetor of BH. Have it intersect AB at M' and BC at N'. M'HN' is congruent to M'BN'. M'N' is parallel to AC so interior angles the same. Futz to show AM'H and HN'C are iscoselles. So AM' = M'H = M'B so 2AM' - AB so M' is midpoint. Same with N'. So M = M' and N=N' and the triangles are congruent. In other words go backward. Assume the triangles are congruent and prove the points most be midpoints are mid points. Sep 7, 2016 at 0:42

(Someone draw a picture of this because I don't know how)

To answer your second question, yes, a segment between two midpoints is parallel to the triangle's opposite side. In this case, call the intersection between $\overline{BH}$ and $\overline{MN}$ $P$, and draw altitude $\overline{MH'}$ perpendicular to $\overline{AC}$. Notice that $\overline{MH'} || \overline{BH}$, so $\angle AMH' = \angle ABH$, and therefore, $\angle BAC = \angle BMN$, so $\overline{MN} || \overline{AC}$.

Additionally, in this case $\overline{MN}$ bisects $\overline{BH}$. This is because $\triangle AMH' \cong \triangle MBP$ via Angle-Side-Angle, so $MH' = BP$. We also have $MH' = PH$, so $\overline{MN}$ does indeed bisect $\overline{BH}$, and it forms right angles because $\overline{MN} || \overline{AC}$.

• I don't see how you established that $\angle BAC = \angle BMN$.
– Jeff
Aug 25, 2014 at 23:20
• $\overline{MH'}$ is perpendicular to $\overline{AC}$, so $\angle MH'A = \angle BHA$. We have $\angle AMH′= \angle ABH$ by parallel lines. Logically, that means $\angle BAC= \angle BMN$ (see triangles $\triangle AMH', \triangle MBP$) Aug 25, 2014 at 23:56
• I think that's only true if you've already established that $MN || AC$. But I've found that can be established using a parallelogram with side $CE$ parallel to $AM$.
– Jeff
Aug 26, 2014 at 3:55 By midpoint theorem, MN//AC.

By intercept theorem, BP = PH

Also angle BPM = angle MPH = 90 degrees.

MP = PM (common side)

Thus ⊿BPM is congruent to ⊿HPM (SAS)

Similarly, ⊿BPN is congruent to ⊿HPN

Hence, ⊿BMN is congruent to ⊿HMN.

Note (1) My P is your X.

Note (2) The way you think is essentially correct. I just add in the details with corresponding reasons as support.

Note (3) We (including others) were confused at first and I finally realized that those questions you are asking are in fact your attempt to the question.

• Can you explain how BP=PH? I don't see how the intercept theorem applys. We can only use congruent triangle parallel line theorems.
– Jeff
Aug 26, 2014 at 14:33
• @Jeff From BM = MA (given); and MP(N) // AH(C) (proved), BP = PH (intercept theorem).
– Mick
Aug 27, 2014 at 3:38
• @Jeff Intercept theorem follows right after, congruent triangles, parallel theorem and the midpoint theorem. It is not that advanced and not that difficult to prove and learn.
– Mick
Aug 27, 2014 at 3:43

The circles with $AB$ and $BC$ as diameters have $BH$ as their radical axis and $M,N$ as the respective centers. Its clear now that $MB= MH, NB = NH$ and congruence follows