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: $\operatorname{ar}(BYXD) = \operatorname{ar}(ABMN)$.

(N/A) To prove: $\operatorname{ar}(BYXD) = \operatorname{ar}(ABMN)$.
$1$. Consider $\Delta MBC$ and $\Delta ABD$. We have $MB = AB$ (sides of square $ABMN$) and $BC = BD$ (sides of square $BCED$). Also,$\angle MBC = \angle MBA + \angle ABC = 90^\circ + \angle ABC$ and $\angle ABD = \angle ABC + \angle CBD = \angle ABC + 90^\circ$. Thus,$\angle MBC = \angle ABD$.
$2$. By $SAS$ congruence,$\Delta MBC \cong \Delta ABD$. Therefore,$\operatorname{ar}(\Delta MBC) = \operatorname{ar}(\Delta ABD)$.
$3$. Since $\Delta ABD$ and rectangle $BYXD$ are on the same base $BD$ and between the same parallels $AX$ and $BD$,$\operatorname{ar}(BYXD) = 2 \operatorname{ar}(\Delta ABD)$.
$4$. Since square $ABMN$ and $\Delta MBC$ are on the same base $MB$ and between the same parallels $MB$ and $NC$,$\operatorname{ar}(ABMN) = 2 \operatorname{ar}(\Delta MBC)$.
$5$. Since $\operatorname{ar}(\Delta ABD) = \operatorname{ar}(\Delta MBC)$,it follows that $2 \operatorname{ar}(\Delta ABD) = 2 \operatorname{ar}(\Delta MBC)$.
$6$. Therefore,$\operatorname{ar}(BYXD) = \operatorname{ar}(ABMN)$.