In the 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)△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(AMBN)+ar(ACFG)
Given : In△ABC,∠A=90∘
BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively
AX⊥DE meeting DE at X
To Prove:
(i) In △MBCand△ABD,MB=ABBC=BD∠MBC=∠ABD∴MBC≅△ABD(ii)∵△ABD and rectangle BYXD are on eh same base BD and between the same parallels
∴ar(△ABD)=12ar(rect.BYXD)⇒ar(rect.BYXD)=2ar(△ABD)⇒ar(rect.BYXD)=2ar(△MBC)
(iii) Similarly, △MBC and MBAN are on he same base MB and between the same parallels
∴ar(△MBC)=ar(sq.ABMN)
From (ii) and (iii)
ar(sq. ABMN ) = ar(rect. BYXD)
In △FCBand△ACE,FC=ACCB=CE∠FCB≅∠ACE∴△FCB≅△ACE
(v) ∵△FCB≅△ACE∴ar(△FCB)=ar(△ACE)∵△ACE and rectangle CYXE are on the same base and between he same parallels.
∴2ar(△ACE)=ar(CYXC)⇒2ar(△FCB)=ar(CYXE)
(vi) △FCB and rectangle FCAG are on eh base FC and between the same parallels
∴2ar(△FCB)=ar(FCAG)
From (iv) and (vi)
BC2=AB2+AC2⇒BC×BD=AB×MB+AC×FC⇒ar(BCED)=ar(ABMN)+ar(ACFG)