Data : ABC is a right angled triangled at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y.
To Prove:
(i) ∆MBC ≅ ∆ABD
(ii) ar.(BYXD) = 2ar.(∆MBC)
(iii) ar.(BYXD) = ar.(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar.(CYXE) = 2ar.(∆FCB)
(vi) ar.(CYXE) = ar.(ACFG)
(vii) ar.(BCED) = ar.(ABMN) + ar.(ACFG).
Proof: (i) Each angle of square is 90°.
∴ ∠ABM = ∠DBC = 90°
∠ABM + ∠ABC = ∠DBC + ∠ABC
∴ ∠MBC = ∠ABD
In ∆MBC and ∆ABD,
∠MBC = ∠ABD (Proved)
MB = AB BC = AD
∴ ∆MBC ≅ ∆ABD (ASA Postulate)
(ii) ∆MBC = ∆ABD
∴ ar.(∆MBC) = ar.(∆ABD) ………… (i)
AX ⊥ DE, BD ⊥ DE
∴ BD || AX
∆ABD and quadrilateral BYXD are on base BD and in between BD || AX.
ar.(∆ABD) = \(\frac{1}{2}\)× ar.(BYXD)
BYXD = 2 × ar.(∆ABC)
BYXD = 2 × ar.(∆MBC) …………(ii)
(iii) ∆MBC, quadrilateral ABMN are on base MB and in between MB || NC.
∴ ar.(∆MBC) = \(\frac{1}{2}\) × ar.(ABMN)
2 × ar.(∆MBC) = ar.(ABMN)
ar.(BYXD) = ar.(ABMN) ……………….. (iii)
(iv) Each angle of square is 90°.
∴ ∠FCA = ∠BCE = 90°
∠FCA + ∠ACB = ∠BCE + ∠ACB
∴ ∠FCB = ∠ACE
In ∆FCB and ∆ACE,
∠FCB = ∠ACE
FC = AC
CB = CE
∴ ∆FCB ≅ ∆ACE (SAS postulte)
(v) AX ⊥ DE, CE ⊥ DE (Data)
∴ CE || AX
∆ACE, quadrilateral CYXE are on base CE and in between CE || AX.
∴ ar.(∆ACE) = \(\frac{1}{2}\)ar.(CYXE)
ar.(CYXE) = 2 × ∆ACE …………… (iv)
∆FCB ≅ ∆ACE Proved.
∴ W(∆FCB) = W(∆ACE) …………… (v)
Comparing (iv) and (v),
ar.(CYXE) = 2 × ar.(∆FCB) ……………. (vi)
(vi) ∆FCB, and quadrilateral ACFG are on base CF and in between CF || BG.
∴ ar.(∆FCB) = \(\frac{1}{2}\) × ar.(∆CFG)
ar.(ACFG) = 2 × ar.(∆FCB) …………… (vii)
(vii) ar.(BCED) = ar.(BYXD) + ar.(CYXE)
∴ 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.)
