In the figure,$ABC$ is a right-angled triangle with the right angle at $A$. $BCED$,$ACFG$,and $ABMN$ are squares constructed on the sides $BC$,$CA$,and $AB$ respectively. The line segment $AX \perp DE$ meets $BC$ at $Y$. Show that: $\Delta FCB \cong \Delta ACE$.
(N/A) To prove: $\Delta FCB \cong \Delta ACE$
In $\Delta FCB$ and $\Delta ACE$:
$1$. $FC = AC$ (Sides of the same square $ACFG$)
$2$. $CB = CE$ (Sides of the same square $BCED$)
$3$. $\angle FCA = \angle BCE = 90^\circ$ (Angles of squares)
Adding $\angle ACB$ to both sides:
$\angle FCA + \angle ACB = \angle BCE + \angle ACB$
$\angle FCB = \angle ACE$
Therefore,by the $SAS$ (Side-Angle-Side) congruence criterion:
$\Delta FCB \cong \Delta ACE$
In $\Delta FCB$ and $\Delta ACE$:
$1$. $FC = AC$ (Sides of the same square $ACFG$)
$2$. $CB = CE$ (Sides of the same square $BCED$)
$3$. $\angle FCA = \angle BCE = 90^\circ$ (Angles of squares)
Adding $\angle ACB$ to both sides:
$\angle FCA + \angle ACB = \angle BCE + \angle ACB$
$\angle FCB = \angle ACE$
Therefore,by the $SAS$ (Side-Angle-Side) congruence criterion:
$\Delta FCB \cong \Delta ACE$
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