AB and CD are parller. E is any point inside the line EF and EG are the perpendicular on AD and CD respectively lets prove that E,F and G are colliner
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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.
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