Question

in the given figure ABCD is a trapezium in which ab parallel to dc and e is the midpoint of the line segment EF which is parallel to AC intersect BC at F show that f is the midpoint of BC

Answer 1

AD = BC ...SIDES OF TRAPEZIUM

AD = AE + ED

AE + ED = BC.....1

AB|| DC.....GIVEN ....2

EF || AB......GIVEN ..3

FROM 2 AND 3.....DC || EF

AB || EF || DC

THEREFORE

Distance between AE = BF

distance between DE = CF

substituing in equation 1

AE + ED = BC

BF + CF = BC

therefore, F is midpoint of BC

Hence proved

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