In the following figure, ABC is a right triangle right angled at A.

Solution:

(i) We know that each angle of a square is $90^{\circ}$.

Hence, $\angle \mathrm{ABM}=\angle \mathrm{DBC}=90^{\circ}$

$\Rightarrow \angle A B M+\angle A B C=\angle D B C+\angle A B C$

$\Rightarrow \angle \mathrm{MBC}=\angle \mathrm{ABD}$

In $\triangle \mathrm{MBC}$ and $\triangle \mathrm{ABD}$,

$\angle \mathrm{MBC}=\angle \mathrm{ABD}($ Proved above $)$

$\mathrm{MB}=\mathrm{AB}$ (Sides of square ABMN)

$B C=B D($ Sides of square BCED $)$

$\therefore \triangle \mathrm{MBC} \cong \triangle \mathrm{ABD}(\mathrm{SAS}$ congruence rule)

(ii) We have

$\triangle \mathrm{MBC} \cong \triangle \mathrm{ABD}$

$\Rightarrow \operatorname{ar}(\Delta \mathrm{MBC})=\operatorname{ar}(\triangle \mathrm{ABD}) \ldots(1)$

It is given that $A X \perp D E$ and $B D \perp D E$ (Adjacent sides of square BDEC)

$\Rightarrow \mathrm{BD} \| \mathrm{AX}$ (Two lines perpendicular to same line are parallel to each other)

$\triangle A B D$ and parallelogram $B Y X D$ are on the same base $B D$ and between the same parallels $B D$ and $A X$.

$\therefore \operatorname{ar}(\triangle \mathrm{ABD})=\frac{1}{2} \operatorname{ar}(\mathrm{BYXD})$

$\operatorname{ar}(B Y X D)=2 \operatorname{ar}(\triangle A B D)$

Area $(B Y X D)=2$ area $(\Delta M B C)[U \operatorname{sing}$ equation $(1)] \ldots(2)$

(iii) $\triangle \mathrm{MBC}$ and parallelogram $\mathrm{ABMN}$ are lying on the same base $\mathrm{MB}$ and between same parallels $\mathrm{MB}$ and $\mathrm{NC}$.

$\therefore \operatorname{ar}(\Delta \mathrm{MBC})=\frac{1}{2} \operatorname{ar}(\mathrm{ABMN})$

$2 \operatorname{ar}(\Delta \mathrm{MBC})=\operatorname{ar}(\mathrm{ABMN})$

$\operatorname{ar}(B Y X D)=\operatorname{ar}(A B M N)[$ Using equation $(2)] \ldots(3)$

(iv) We know that each angle of a square is $90^{\circ}$.

$\therefore \angle \mathrm{FCA}=\angle \mathrm{BCE}=90^{\circ}$

$\Rightarrow \angle \mathrm{FCA}+\angle \mathrm{ACB}=\angle \mathrm{BCE}+\angle \mathrm{ACB}$

$\Rightarrow \angle \mathrm{FCB}=\angle \mathrm{ACE}$

In $\triangle \mathrm{FCB}$ and $\triangle \mathrm{ACE}$,

$\angle F C B=\angle A C E$

$\mathrm{FC}=\mathrm{AC}($ Sides of square $\mathrm{ACF} \mathrm{G})$

$C B=C E($ Sides of square BCED $)$

$\triangle \mathrm{FCB} \cong \triangle \mathrm{ACE}(S A S$ congruence rule $)$

(v) It is given that $A X \perp D E$ and $C E \perp D E$ (Adjacent sides of square BDEC)

Hence, CE II AX (Two lines perpendicular to the same line are parallel to each other)

Consider $\triangle A C E$ and parallelogram $C Y X E$

$\triangle A C E$ and parallelogram $C Y X E$ are on the same base $C E$ and between the same parallels $C E$ and $A X$.

$\therefore \operatorname{ar}(\Delta \mathrm{ACE})=\frac{1}{2} \operatorname{ar}(\mathrm{CYXE})$

$\Rightarrow \operatorname{ar}(C Y X E)=2 \operatorname{ar}(\triangle A C E) \ldots(4)$

We had proved that

$\therefore \triangle \mathrm{FCB} \cong \triangle \mathrm{ACE}$..(5)

On comparing equations (4) and (5), we obtain

$\operatorname{ar}(C Y X E)=2 \operatorname{ar}(\Delta F C B) \ldots(6)$

(vi) Consider $\triangle \mathrm{FCB}$ and parallelogram $\mathrm{ACFG}$

$\triangle \mathrm{FCB}$ and parallelogram $\mathrm{ACFG}$ are lying on the same base $\mathrm{CF}$ and between the same parallels $\mathrm{CF}$ and $\mathrm{BG}$.

$\therefore \operatorname{ar}(\Delta \mathrm{FCB})=\frac{1}{2} \operatorname{ar}(\mathrm{ACFG})$

$\Rightarrow \operatorname{ar}(A C F G)=2 \operatorname{ar}(\Delta F C B)$

$\Rightarrow \operatorname{ar}($ ACFG $)=\operatorname{ar}($ CYXE $)[$ Using equation $(6)] \ldots(7)$

(vii) From the figure, it is evident that

$\operatorname{ar}(B C E D)=\operatorname{ar}(B Y X D)+\operatorname{ar}(C Y X E)$

$\Rightarrow \operatorname{ar}(B C E D)=\operatorname{ar}(A B M N)+\operatorname{ar}(A C F G)[U \operatorname{sing}$ equations $(3)$ and $(7)]$