(i) We know that each angle of a square is 90°. Hence, ⊥ABM = ⊥DBC = 90
⊥ ABM + ⊥ABC = ⊥DBC + ⊥ABC
⊥ MBC = ⊥ABD
⊥MBC = ⊥ABD (Proved above)
MB = AB ( Sides of square ABMN)
BC = BD ( Sides of square BCED)
⊥ ΔMBC ⊥ ΔABD ( SAS congruence rule)
(ii) We have
ΔMBC ⊥ ΔABD
⊥ ar (ΔMBC) = ar (ΔABD) ... (1)
It is given that AX ⊥ DE and BD ⊥ DE (Adjacent sides of square BDEC)
⊥ BD || AX (Two lines perpendicular to same line are parallel to each other) ΔABD and parallelogram BYXD are on the same base BD and between the same parallels BD and AX.
∴ ar (ΔABD) = 1/2 ar (BYXD)
ar (BYXD) = 2 ar (ΔABD)
Area (BYXD) = 2 area (ΔMBC) [Using equation (1)] ... (2)
(iii) ΔMBC and parallelogram ABMN are lying on the same base MB and between same parallels MB and NC.
∴ ar (ΔMBC) = 1/2 ar (ABMN)
2 ar (ΔMBC) = ar (ABMN)
ar (BYXD) = ar (ABMN) [Using equation (2)] ... (3)
(iv) We know that each angle of a square is 90°. ⊥ FCA = ⊥BCE = 90
⊥ FCA + ⊥ACB = ⊥BCE + ⊥ACB ⊥ ⊥FCB = ⊥ACE
In ΔFCB and ΔACE, ⊥FCB = ⊥ACE
FC = AC ( Sides of square ACFG) CB = CE ( Sides of square BCED) ΔFCB ⊥ ΔACE ( SAS congruence rule)
(v) It is given that AX ⊥ DE and CE ⊥ DE (Adjacent sides of square BDEC)
Hence, CE || AX (Two lines perpendicular to the same line are parallel to each other) Consider ΔACE and parallelogram CYXE
ΔACE and parallelogram CYXE are on the same base CE and between the same parallels CE and AX.
∴ ar (ΔACE) = 1/2 ar (CYXE)
⊥ ΔFCB ⊥ ΔACE
ar (ΔFCB) ⊥ ar (ΔACE) ... (5)
On comparing equations (4) and (5), we obtain ar (CYXE) = 2 ar (ΔFCB) ... (6)
(vi) Consider ΔFCB and parallelogram ACFG
ΔFCB and parallelogram ACFG are lying on the same base CF and between the same parallels CF and BG.
∴ ar (ΔFCB) = 1/2 ar (ACFG)
⊥ ar (ACFG) = 2 ar (ΔFCB)
⊥ ar (ACFG) = ar (CYXE) [Using equation (6)] ... (7)
(vii) From the figure, it is evident that ar (BCED) = ar (BYXD) + ar (CYXE)
⊥ ar (BCED) = ar (ABMN) + ar (ACFG) [Using equations (3) and (7)]