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# In 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 DE meets BC at Y. Show that:

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In 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 DE meets BC at Y. Show that:

(i) ΔMBC ≅ ΔABD

(ii) ar(BYXD) = 2ar(MBC)

(iii) ar(BYXD) = ar(ABMN)

(iv) ΔFCB ≅ ΔACE

(v) ar(CYXE) = 2ar(FCB)

(vi) ar(CYXE) = ar(ACFG)

(vii) ar(BCED) = ar(ABMN)+ar(ACFG)

Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.

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(i) We know that each angle of a square is 90°. Hence, ∠ABM = ∠DBC = 90º

∴ ∠ABM + ∠ABC = ∠DBC + ∠ABC

∴ ∠MBC = ∠ABD

In ∆MBC and ∆ABD,

∠MBC = ∠ABD (Proved above)

MB = AB (Sides of square ABMN)

BC = BD (Sides of square BCED)

∴ ∆MBC ≅ ∆ABD (SAS congruency)

(ii) We have

∆MBC ≅ ∆ABD

∴ ar (∆MBC) = ar (∆ABD) … (i)

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)

∆ABD and parallelogram BYXD are on the same base BD and between the same parallels BD and AX.

Area (∆YXD) = 2 Area (∆MBC) [From equation (i)] … (ii)

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

2 ar (∆MBC) = ar (ABMN)

ar (∆YXD) = ar (ABMN) [From equation (ii)] … (iii)

(iv) We know that each angle of a square is 90°.

∴ ∠FCA = ∠BCE = 90º

∴ ∠FCA + ∠ACB = ∠BCE + ∠ACB

∴ ∠FCB = ∠ACE

In ∆FCB and ∆ACE,

∠FCB = ∠ACE

FC = AC (Sides of square ACFG)

CB = CE (Sides of square BCED)

∆FCB ≅ ∆ACE (SAS congruency)

(v) AX ⊥ DE and CE ⊥ DE (Adjacent sides of square BDEC) [given]

Hence,

CE || AX (Two lines perpendicular to the same line are parallel to each other)

Consider BACE and parallelogram CYXE

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

∴ar (∆YXE) = 2ar (∆ACE) … (iv)

∴ ∆FCB ≅ ∆ACE

ar (∆FCB) ≅ ar (∆ACE) … (v)

From equations (iv) and (v), we get

ar (CYXE) = 2 ar (∆FCB) … (vi)

(vi) Consider BFCB and parallelogram ACFG

BFCB and parallelogram ACFG are lying on the same base CF and between the same parallels CF and BG.

∴ ar (ACFG) = 2 ar (∆FCB)

∴ ar (ACFG) = ar (CYXE) [From equation (vi)] …....(vii)

(vii) From the figure, we can observe that

ar (∆CED) = ar (∆YXD) + ar (CYXE)

∴ ar (∆CED) = ar (ABMN) + ar (ACFG) [From equations (iii) and (vii)].

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