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: $\operatorname{ar}(CYXE) = \operatorname{ar}(ACFG)$.

(A) To prove $\operatorname{ar}(CYXE) = \operatorname{ar}(ACFG)$:
$1$. Consider $\Delta ABD$ and $\Delta MBC$. We have $AB = MB$ (sides of square $ABMN$),$BD = BC$ (sides of square $BCED$),and $\angle ABD = \angle ABC + \angle CBD = \angle ABC + 90^\circ = \angle ABC + \angle MBA = \angle MBC$.
$2$. By $SAS$ congruence,$\Delta ABD \cong \Delta MBC$.
$3$. Since $\Delta ABD$ and square $ABMN$ are on the same base $AB$ and between the same parallels,$\operatorname{ar}(\Delta ABD) = \frac{1}{2} \operatorname{ar}(ABMN)$.
$4$. Similarly,$\Delta MBC$ and rectangle $BYXD$ are on the same base $BD$ and between the same parallels,so $\operatorname{ar}(\Delta MBC) = \frac{1}{2} \operatorname{ar}(BYXD)$.
$5$. Since $\Delta ABD \cong \Delta MBC$,their areas are equal. Thus,$\frac{1}{2} \operatorname{ar}(ABMN) = \frac{1}{2} \operatorname{ar}(BYXD)$,which implies $\operatorname{ar}(ABMN) = \operatorname{ar}(BYXD)$.
$6$. By a similar argument,$\operatorname{ar}(ACFG) = \operatorname{ar}(CYXE)$.
$7$. Thus,$\operatorname{ar}(CYXE) = \operatorname{ar}(ACFG)$.