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).
Best Answer
(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 $\overline{MN} || \overline{AC}$.