Question

In the adjoining figure, ABCD is a parallelogram in which E is the mid-point of DC and F is a point on AC such that 1 CF = AC. If EF is produced to meet BC in G, prove that G 4 is the mid-point of BC.


[Hint. Join BD and remember that the diagonals of a Ilgm bisect each other.]


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Answer 1

Answer:

We know that the diagonals of a parallelogram bisect each other.

Therefore, according to midpoint theorem in ∆CSD PQ || DS If PQ || DS

we can say that QR || SB In ∆ CSB, Q is midpoint of CS and QR ‖ SB

Applying converse of midpoint theorem , we conclude that R is the midpoint of CB

This completes the proof.