Question

10. In a trapezium ABCD, AB || DC and point E is the mid-point
of the side AD. A line through the point E and parallel to the
side AB meets the line BC in F. Prove that Fis the mid-point
of BC​

Answer 1

Answer:

HENCE PROOVED

Step-by-step explanation:

Given ABCD is a trapezium in which AB || DC and EF||AB|| CD.

Construction Join, the diagonal AC which intersects EF at O.

To show F is the mid-point of BC.

Proof Now, in ΔADC, E is the mid-point of AD and OE || CD. Thus, by mid-point theorem, O is mid-point of AC.

Now, in ΔCBA, 0 is the mid-point of AC and OF || AB.

So, by mid-point theorem, F is the mid-point of BC.

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Answer 2

Solution:

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

AD/EA = AP/AQ = 2

This gives AP=2AQ

i.e, 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