In the figure,$ABC$ is a right triangle right-angled at $A$. $BCED$,$ACFG$,and $ABMN$ are squares on the sides $BC$,$CA$,and $AB$ respectively. Line segment $AX \perp DE$ meets $BC$ at $Y$. Show that: $\Delta MBC \cong \Delta ABD$.

(N/A) We have a right-angled $\Delta ABC$ such that $BCED$,$ACFG$,and $ABMN$ are squares on its sides $BC$,$CA$,and $AB$ respectively. Line segment $AX \perp DE$ is drawn such that it meets $BC$ at $Y$.
To prove that $\Delta MBC \cong \Delta ABD$:
In $\Delta ABD$ and $\Delta MBC$,we have:
$AB = MB$ (Sides of the square $ABMN$)
$BD = BC$ (Sides of the square $BCED$)
$\angle MBA = 90^\circ$ and $\angle CBD = 90^\circ$ (Angles of squares)
Therefore,$\angle MBA = \angle CBD = 90^\circ$.
Adding $\angle ABC$ to both sides:
$\angle MBA + \angle ABC = \angle CBD + \angle ABC$
$\angle MBC = \angle ABD$
Thus,by the $SAS$ (Side-Angle-Side) congruence criterion:
$\Delta MBC \cong \Delta ABD$.