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: $\operatorname{ar}(BYXD) = 2 \operatorname{ar}(MBC)$

(N/A) To prove: $\operatorname{ar}(BYXD) = 2 \operatorname{ar}(\Delta MBC)$
$1$. Consider $\Delta ABD$ and $\Delta MBC$. In these triangles,$AB = MB$ (sides of square $ABMN$) and $BD = BC$ (sides of square $BCED$). Also,$\angle ABD = \angle ABC + 90^\circ$ and $\angle MBC = \angle ABC + 90^\circ$. Thus,$\angle ABD = \angle MBC$. By $SAS$ congruence,$\Delta ABD \cong \Delta MBC$. Therefore,$\operatorname{ar}(\Delta ABD) = \operatorname{ar}(\Delta MBC)$.
$2$. Parallelogram $BYXD$ and $\Delta ABD$ are on the same base $BD$ and between the same parallels $BD$ and $AX$. Therefore,$\operatorname{ar}(\Delta ABD) = \frac{1}{2} \operatorname{ar}(BYXD)$.
$3$. Substituting the result from step $1$ into step $2$: $\operatorname{ar}(\Delta MBC) = \frac{1}{2} \operatorname{ar}(BYXD)$.
$4$. Hence,$\operatorname{ar}(BYXD) = 2 \operatorname{ar}(\Delta MBC)$.