12. In the given figure, ABCD is a parallelogram. E is the mid-point of CD and P is a point on AC such that PC = 1/4 AC. EP produced meets BC at F. Prove that (i) F is the mid-point ofBC. (ii) 2EF=BD Share with your friends Share 1 Neha Sethi answered this Dear student Given: ABCD is a ∥gm. and E is the mid point of CD.Also, PC=14ACTo prove: F is the mid point of BCConstruction: Join B and D and suppose it cut AC at O.It is given that ABCD is a ∥gm.So, OC=12AC because diagonals of a ∥gm bisect each other.Also, PC=14ACSo, PC=12OCIn △DCO ,E and P are mid points of CD and OC respectively.Theorem states the line segment joining the mid points of any two sides of a triangle is parallel to the third side and equal to half of it.So, we get EP∥DOalso, in △COB, P is the mid point of OC and PF ∥OBSo, F is the mid point of BC Hence proved.ii Since E is the mid point of CD and F is the mid point of BCSo, EF=12BD segment joining the mid points of the two sides of a △ is half of the third side.⇒2EF=BDHence proved Regards 3 View Full Answer