In triangle $ABC$, $BA=BC$ and $\angle B=90^\circ$. $D,E$ are the points on $AB,BC$ respectively such that $AD=CE$. $M,N$ are points on $AC$ such that $DM$ is perpendicular to $AE$ and $BN$ is perpendicular to $AE$. Prove that $MN=NC$.
How can we solve this geometric problem by application of midpoint theorem?
Midpoint theorem states that in triangle ABC if D,E are midpoints of AB,AC respectively then AB is parallel to BC and AB =BC/2.
And converse of midpoint theorem states that:- In triangle ABC if D is the midpoint of AB and a line say U is drawn passing through D parallel to BC then U intersects AC(Say at E) and A-E-C and AE=EC.
Yes, you can use this theorem, but you need some additional artwork here. Please excuse my hand drawing, as I still do not have an appropriate package at hand.
You need to continue the line containing $BC$ beyond $B$ till the intersection with $DM$. Call the point of intersection $P$ (I chose to connect $A$ and $P$, though this is not strictly necessary).
Now $\angle PDB = \angle AEB$ as the angles whose sides are mutually perpendicular. $\angle PBD = \angle ABE = \pi/2$. $\Delta ABC$ is isosceles, hence $AB=BC$. $AD=EC$ by assumption, therefore $BD=BE$. Thus $\Delta AEB= \Delta PDB$.
Therefore, $PB=AB=BC$. $PM\parallel BN$ since they are both $\perp$ $AE$. Now you can invoke the converse midpoint theorem for $\Delta PMC$ and line $NB$.