Question

In the adjoining figure, ABCD is a trapezium in which AB || DC and E
is the mid-point of AD. If EF || AB meets BC at F, show that F is the mid-point of BC
(Hint. AB || EF || DC and AD is the transversal such that AE = ED.
Another transversal cuts them at B, F and C respectively.
So, BF = FC:]

please friends at least give me this answer please. ​it is argent .

Answer 1

Answer:

GIVEN : ABCD IS A TRAPEZIUM WHERE AB║DC AND E IS MIDPOINT OF AD

SO AE = DE AND EF║AB

TO PROVE : F IS MIDPOINT OF BC i.e., BF = CF

PROOF : LET EF INTERSECT DB AT G

IN Δ ABD

E IS MIDPOINT OF AD

AND EG║AB (as EF║AB, parts of parallel lines are parallel)

∴G WILL BE MIDPOINT OF DB (line drawn through midpoint of 1 side of triangle, parallel to another side bisects the third side.

GIVEN : EF║AB AND AB║CD

∴ EF║CD

IN Δ BCD,

G IS MIDPOINT OF SIDE BD

AND GF║CD ( as EF║CD, parts of parallel lines are parallel )

∴  F is the mid-point of BC (line drawn through midpoint of 1 side of triangle, parallel to another side bisects the third side.)

···HENCE PROVED···

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Answer 2

Answer:

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Step-by-step explanation:

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