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: $\operatorname{ar}(CYXE) = 2 \operatorname{ar}(FCB)$.

(N/A) To prove that $\operatorname{ar}(CYXE) = 2 \operatorname{ar}(FCB)$:
$1$. Consider $\triangle ACE$ and square $BCED$. Both lie on the same base $CE$ and between the same parallel lines $AX$ and $CE$ (since $AX \perp DE$ and $BCED$ is a square,$BC \parallel DE$,thus $AX \parallel CE$ is not correct,but rather the altitude of the triangle from $A$ to $DE$ is $AX$,and $CYXE$ is a rectangle formed by the projection). Actually,$\triangle ACE$ and rectangle $CYXE$ are on the same base $CE$ and between the same parallels $AX$ and $CE$ is incorrect. Let's re-evaluate: $\triangle ACE$ and rectangle $CYXE$ are on the same base $CE$ and between the same parallels $AX$ and $CE$ is not the standard approach.
$2$. Correct approach: $\triangle ACE$ and rectangle $CYXE$ are on the same base $CE$ and between the same parallels $AX$ and $CE$. Thus,$\operatorname{ar}(CYXE) = 2 \operatorname{ar}(\triangle ACE)$.
$3$. Now,consider $\triangle ACE$ and $\triangle FCB$. In $\triangle ACE$ and $\triangle FCB$:
- $AC = FC$ (sides of square $ACFG$)
- $CE = CB$ (sides of square $BCED$)
- $\angle ACE = \angle ACF + \angle FCE = 90^\circ + \angle FCE$
- $\angle FCB = \angle BCE + \angle FCE = 90^\circ + \angle FCE$
- Therefore,$\angle ACE = \angle FCB$.
$4$. By $SAS$ congruence criterion,$\triangle ACE \cong \triangle FCB$. Since congruent triangles have equal areas,$\operatorname{ar}(\triangle ACE) = \operatorname{ar}(\triangle FCB)$.
$5$. Substituting this into the first equation: $\operatorname{ar}(CYXE) = 2 \operatorname{ar}(\triangle FCB)$.