Question

ABCD is a trapezium in which AB║DC, BD is a diagonal

and E is the mid-point of AD. A line is drawn through E

parallel to AB intersecting BC at F (see figure ).

Show that F is the mid-point of BC.​

Answer 1

you can prove it by mid and converse of mid point theorem

Given ABCD is a trapezium.

We have to prove, F is the mid point of BC, i.e., BF=CF

Let EF intersect DB at G.

In ΔABD E is the mid point of AD and EG∣∣AB.

∴ G will be the mid-point of DB.

Now EF∣∣AB and AB∣∣CD

∴ EF∣∣CD

∴ In ΔBCD, GF∣∣CD

⇒ F is the mid point of BC.

Answer 2

                                                $\huge\red{{\fbox{\tt{Answer}}$

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Trapezium:

A quadrilateral in which one pair of opposite sides are parallel is called a trapezium.

Converse of mid-point theorem:

The line drawn through the midpoint of one side of a triangle, parallel to another side bisects the third side.

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Given:  ABCD is a trapezium where AB ∥ DC E is the mid-point of AD, i.e., AE = DE & EF ∥ AB

To prove:  F is mid-point of BC, i.e., BF = CF

Proof: Let EF intersect DB at G.

In ∆ ABD

E is the mid-point of AD

EG ∥ AB

∴ G is the mid-point of DB $(converse\ of\ mid-point\ theorem)$

Given EF ∥ AB and AB ∥ CD,

∴ EF ∥ CD

In ΔBCD

G is the mid-point of BD

GF ∥ CD

∴ F is the mid-point of BC $(converse\ of\ mid-point\ theorem)$

Hence proved

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