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.