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: $\operatorname{ar}(BCED) = \operatorname{ar}(ABMN) + \operatorname{ar}(ACFG)$

(N/A) To prove: $\operatorname{ar}(BCED) = \operatorname{ar}(ABMN) + \operatorname{ar}(ACFG)$.
$1$. Join $AD$ and $FC$. Observe $\triangle ABD$ and $\triangle MBC$. We have $AB = MB$ (sides of square $ABMN$),$BD = BC$ (sides of square $BCED$),and $\angle ABD = \angle ABC + 90^\circ = \angle MBC + 90^\circ = \angle MBC$. Thus,$\triangle ABD \cong \triangle MBC$ by $SAS$ congruence.
$2$. Since $\triangle ABD$ and square $ABMN$ are on the same base $AB$ and between the same parallels $AB$ and $MD$,$\operatorname{ar}(ABD) = \frac{1}{2} \operatorname{ar}(ABMN)$.
$3$. Similarly,$\triangle MBC$ and rectangle $BYXD$ are on the same base $BD$ and between the same parallels $BD$ and $CX$,so $\operatorname{ar}(MBC) = \frac{1}{2} \operatorname{ar}(BYXD)$.
$4$. Since $\triangle ABD \cong \triangle MBC$,their areas are equal. Therefore,$\frac{1}{2} \operatorname{ar}(ABMN) = \frac{1}{2} \operatorname{ar}(BYXD)$,which implies $\operatorname{ar}(ABMN) = \operatorname{ar}(BYXD)$.
$5$. By a similar argument,$\operatorname{ar}(ACFG) = \operatorname{ar}(CYXE)$.
$6$. Adding these,$\operatorname{ar}(ABMN) + \operatorname{ar}(ACFG) = \operatorname{ar}(BYXD) + \operatorname{ar}(CYXE) = \operatorname{ar}(BCED)$.