# ABCD is trapezium in which AB||DC. E is the mid point of AD. If F is point on side BC such that segment EF is parallel to side DC, show that EF = ½ (AB + CD).

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by pravin3103

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by pravin3103

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The Brainliest Answer!

EF is parallel to side DC. Then we have AB||DC||EF

Hence we have also trapezium ABFE and trapezium EFCD

Let AP be the perpendicular to DC and this intersects EF at Q.

AQ will be perpendicular to EF.

For triangle APD and AQE we have AD/EA=AP/AQ=2

This gives AP=2AQ ie AQ = QP

Consider the area we have area ABCD= area ABFE + area EFCD (1/2)AP*(AB+DC)= (1/2)AQ*(AB+EF) + (1/2)QP*(EF+DC)

⇒ AP(AB+DC)= AP*AB/2 + AP*EF/2 +AP*EF/2 +AP*DC/2

⇒ AP*AB/2 + AP*DC/2 = AP*EF/2 + AP*EF/2

⇒ AP*(AB+DC)/2 = AP*EF

⇒ EF=(AB+DC)/2