The correct option is
B ABBC=DEEFGiven: l||m||n and we have two transversal p and q, there cut A,B,C and D,E,F.
By Construction, we take point O at the intersection of both transversed line p and q.
and FX⊥OCandCY⊥OF.Join AF,BF,DC and EC.
Since,FX⊥OC,SoFXis height of△OCF&△BCF.
ar(△OCF)=12×OC×FX
ar(△BCF)=12×BC×FX
∴ar(△BCF)ar(△OCF)=BCOC−(1)
Similarily,Since CY⊥OF,So CY is height of△OCF&△ECF,
∴ar(ECF)ar(OCF)=EFOF−(2)
But△BCFand△ECF are on the same base CF b/w same parallel straight lines BC& CF {m||n}
△BCF=△ECF−(3)
∴BCOC=EFOF−(A)
Since,FX⊥OC,FX is height of△OCF&△ACF
ar(△OCF)=12×OC×FX
ar(△ACF)=12×AC×FX
ar(ACF)ar(OCF)=ACOC=AB+BCOC−(4)
Since,CY⊥OF,SoCYisheightof△OCF&△DCF.
ar(DCF)ar(OCF)=DE+EFOF
But△ACF&△DCF are on same base CF and b/w same parallel straight line,
Since,ADandCF{AD||CFasl||n}
△ACF=△DCF−(6)
ABOC+BCOC=DEOF+EFOF
⟹ABOC=DEOF−(B){∵EFOF=BCOC}
∴ABOCBCOC=DEOFEFOF
∴ABBC=DEEF