In the given figure,$AM$ and $BN$ are both perpendicular to $AB$. $MN$ intersects $AB$ at $P$. Also,$P$ is the midpoint of $AB$. Prove that $AM = BN$ and $P$ is the midpoint of $MN$.

(N/A) Consider $\triangle APM$ and $\triangle BPN$.
$1$. $\angle MAP = \angle NBP = 90^{\circ}$ (Given that $AM \perp AB$ and $BN \perp AB$).
$2$. $AP = BP$ ($P$ is the midpoint of $AB$,given).
$3$. $\angle APM = \angle BPN$ (Vertically opposite angles).
Therefore,by the $ASA$ (Angle-Side-Angle) congruence criterion,$\triangle APM \cong \triangle BPN$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$:
- $AM = BN$ (Hence proved).
- $PM = PN$ (Since $PM$ and $PN$ are corresponding sides).
Since $PM = PN$,$P$ is the midpoint of $MN$ (Hence proved).