Question

in a trapezium ABCD in which AB is parallel to CD and DC = 2AB, and EF is drawn parallel to AB such that BE by BC =3/4 prove that 7EF=10AB...and BD intersects EF at G

Answer 1

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ABCD is a trapezium. where AB//CD. and 

DC = 2 AB .......(1) and BE /EC = 3/4.............(2)

from eq(2):

BE/EC = 3/4

BE/EC +1 = 3/4 +1

(BE+EC)/EC = (3+4)/4

BC/EC = 7/4

the triangles CGE and triangles CAB are similar triangles.

therefore ratio of corresponding sides are equal.

EG/AB = EC/BC

EG/AB = 4/7

=> 7×EG=4AB _____(3)

 

since AB//EF//CD therefore

AF/FD = BE/EC = 3/4

FD/AF = 4/3

FD/AF + 1 = 4/3 +1

(FD+AF)/AF = 7/3

AD/AF = 7/3

the triangles AFG and ADC are similar triangles.

therefore => FG/CD = AF/AD

FG/CD = 3/7

=> 7×FG = 3×CD

from eq(1):

7×FG = 3×2AB

7FG = 6AB_______(4)

adding eq(3) and eq(4):

7EG + 7FG = 4AB + 6AB

7(EG+FG) = 10AB

7EF = 10AB __________proved

which is the required result.

hope this helps you.

Devil_king ▄︻̷̿┻̿═━一