  Question

# In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). The line segments AF and EC _____ the diagonal BD. TrisectBisectSimply intersect at two pointsNone of these

Solution

## The correct option is A TrisectGiven: ABCD is a parallelogram and E, F are the mid-points of sides AB and CD respectively. To Prove: Line segments AF and EC trisect the diagonal BD. Proof: Since, ABCD is a parallelogram, AB || DC And, AB = DC (Opposite sides of a parallelogram) ⇒ AE || FC and 12 AB = 12 DC ⇒ AE || FC and AE = FC ∴ AECF is a parallelogram ∴ AF || EC ⇒ EQ || AP and FP || CQ In Δ BAP, E is the mid-point of AB and EQ || AP, so Q is the mid - point of BP. (By converse of mid-point theorem) BQ = OQ ........(i) Again, in  Δ DQC,F is the mid-point of DC and FP || CQ, so P is the mid - point of DQ. (By converse of mid-point theorem) QP = DP  ….(ii) From Equations (i) and (ii) , we get BQ = PQ = PD Hence, CE and AF trisect the diagonal BD.  Suggest corrections   