Question

In the given figure AB parallelFD,AC parallel GE and BD =CE prove that BG= DF and CF=EG

Answer 1

Answer:

HI

PLZ LOOK INTO THE FIGURE TOO FOR REFERENCE!

IN THE FIGURE AB║ FD,

∠ABC =∠FDC

ALSO AC║GE

∠ACB = ∠GEB

CONSIDER THE TWO TRIANGLES

ΔGBE AND ΔFDC

ANGLE B = ANGLE D

ANGLE C = ANGLE E

BD = CE

BD+DE = CE + DE

BE =DC

ΔGBE ≅ΔFDC (A.S.A)

GB/ FD = BE /DC = GE/FC

BUT BE = DC

=BE/DC

=BE/BE

=1

AS GB= FD

GB/FD = BE/DC = 1

GE/FC = BE/DC =1

GE=FC

HENCE PROVED

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