Question

ABCD is a trapezium, in which AB is parallel to DC. E is the mid-point of AD. EF is drawn parallel to DC. Prove that EF: AB+DC/2​

Answer 1

ABCD is trapezium in which AB∥DC.

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 △APD and △AQE we have EAAD=AQAP=2

This gives AP=2AQ  

i.e, AQ=QP

Consider the area  we have area ABCD=area ABFE+ area EFCD                    

(21)AP×(AB+DC)=(21)AQ×(AB+EF)+(21)QP×(EF+DC)

⇒AP(AB+DC)=AP×2AB+AP×2EF+AP×2EF+AP×2DC        

 ⇒AP×2

⇒AP×2AB+AP×2DC=AP×2EF+AP×2EF

⇒AP×2(AB+DC)=AP×EF

EF=2(AB+DC)/2

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