E and F are mid-points of sides AB and CD, respectively of a parallelogram ABCD. AF and CE intersect diagonal BD in P and Q, respectively. Prove that diagonal BD is trisected at P and Q.

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Solution

$Given : In a parallelogram ABCD, E and F are mid - points of sides AB and CD, respectively . To prove : Diagonal BD is trisected at P and Q . Proof : As, E and F are mid - points of sides AB and CD, respectively . \Rightarrow AE = \frac{1}{2} AB and CF = \frac{1}{2} CD But AB = CD and AB ∥ CD (Opposite sides of parallelogram ABCD) \Rightarrow AE = CF and AE ∥ CF ∴ AECF is a parallelogram . \Rightarrow AF ∥ EC . . . . . (i) Now, in ∆ ABP, As, E is the mid - point of AB and QE ∥ AP [From (i)] So, BQ = PQ . . . . . (ii) (Using converse of mid - point theorem) Similarly, in ∆ DQC, As, F is the mid - point of DC and PF ∥ QC [From (i)] So, DP = PQ . . . . . (iii) (Using converse of mid - point theorem) From (ii) and (iii), we get DP = PQ = BQ ∴ Diagonal BD is trisected at P and Q .$

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