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Question

In trapezium ABCD, AB||DA and DC=2AB. EF drawn parallel to AB cuts AD in F

and BC in E such that BEEC=34. Diagonal DB intersects EF at G. Prove that 7FE=10 AB.

1008966_14a2ed3b2305478fb157220d760f7851.png

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Solution

In DFG and DAB, we have

1=2 [AB||DC||EF1 and 2 are corresponding angles]

FDG=ADB [Common]

So, by AA-criterion of similarity, we have

DFGDAB

DFDA=FGAB.......(i)

In trapezium ABCD, we have

EF||AB||DC

AFDF=BEEC

AFDF=34
[BEEC=34(Given)]

AFDF+1=34+1 [Adding 1 on both sides]

AF+DFDF=74

ADDF=74

DFAD=47......(ii)

From (i) and (ii), we get

FGAB=47

FG=47AB........(iii)

In BEG and BCD, we have

BEG=BCD [Corresponding angles]

B=B [Common]

BEGBCD [By AA-criterion of similarity]

BEBC=EGCD

37=EGCD [BEEC=34ECBE=43ECBE+1=43+1BCBE=73]

EG=37CD

EG=37×2AB

EG=67×AB

Adding (iii) and (iv), we get

FG+EG=47AB+67AB

EF=107AB

7EF=10AB [Hence proved]


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