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).
 A: (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 $\overline{MN} || \overline{AC}$.
A: 
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.
A: 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
