# In Fig. 9.34, ABC

(i) Since, each angle of a square is 900

Hence, angle ABM = angle DBC = 900

ABM + ABC = DBC + ABC

MBC = ABD (1)

In ΔMBC and ΔABD,

Angle MBC = Angle ABD (From 1)

MB = AB (Sides of square)

BC = BD (Sides of square)

Therefore,

By SAS congruence rule, (ii) Now, we have: And we know that adjacent sides of square are perpendicular to each other. Hence,

BD perpendicular DE

And,

AX perpendicular DE

Also,

We know that two lines perpendicular to the same line are parallel to each other.

Therefore,

BD ‖ AX

Now, = ( and ‖gm BXYD are on same base BD and between the same parallels BD and AX)

Area (BXYD) = 2 area (ΔABD)

Area (BXYD) = 2 area (ΔMBC) (2)

(iii) ΔMBC and parallelogram ABMN are on same base MB and between the same parallels MB and NC.

Therefore,

Area (ΔMBC) = � ar (ABMN)

2 Area (ΔMBC) = Area (ABMN) = Area(BXYD) (From 2) (3)

(iv) Since, each angle of square is 900

Therefore,

Angle FCA = Angle BCE = 900

FCA + ACB = BCE + ACB

Angle FCB = Angle ACE

Now,

In ΔFCB and ΔACE,

Angle FCB = Angle ACE

FC = AC (Sides of square)

CB = CE (Sides of square)

Therefore,

By SAS congruence, (v) It is given that AX is perpendicular to DE and CE is perpendicular to DE

and,

We know that two lines perpendicular to the same line are parallel to each other.

Hence,

CE ‖ AX

Now,

ΔACE and parallelogram CYXE are on same base CE and between the same parallels CE and AX.

Therefore,

Area (ΔACE) = 1/2 Area (CYXE)

Area (CYXE) = 2 Area (ΔACE) (4)

ΔFCB ΔACE

Hence,

Area (ΔACE) =Area (ΔFCB) (5)

Comparing 4 and 5, we get:

CYXE = 2 Area (ΔFCB) (6)

(vi) ΔFCB and parallelogram ACFG are on same base CF and between the same parallels CF and BG

Hence,

Area (ΔFCB) = � Area (ACFG)

Area (ACFG) = 2 Area (ΔFCB) = Area (CYXE) (Using 6) (7)

(vii) From figure:

Area (BCED) = Area (BXYD) + Area (CYXE)

Area (BCED) = Area (ABMN) + Area (ACFG) (using 3 and 7)

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