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Question

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 AXDE meets BC at Y Show that
(i)MBCABD(ii)ar(BYXD)=2ar(MBC)(iii)ar(BYXD)=ar(ABMN)(iv)FCBACE(v)ar(CYXE)=2ar(FCB)(vi)ar(CYXE)=ar(ACFG)(vii)ar(BCED)=ar(AMBN)+ar(ACFG)

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Solution

Given : InABC,A=90
BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively
AXDE meeting DE at X
To Prove:

(i) In MBCandABD,MB=ABBC=BDMBC=ABDMBCABD(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 FCBandACE,FC=ACCB=CEFCBACEFCBACE
(v) FCBACEar(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+AC2BC×BD=AB×MB+AC×FCar(BCED)=ar(ABMN)+ar(ACFG)


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