Quadrilaterals — Question 5

Given :

ABCD is a parallelogram, where E and F are the mid-points of sides AB and CD.

We know that,

Opposite sides of a parallelogram are equal and parallel.

∴ AB || CD and AB = CD

∴ AE || FC

Since, AB = CD

Dividing both sides of equation by 2, we get :

⇒ $\dfrac{1}{2}AB$ = $\dfrac{1}{2}CD$

⇒ AE = FC

∴ AEFC is a parallelogram.

∴ AF || CE

From figure,

⇒ PF || QC

⇒ EQ || AP

In △ DQC,

F is the mid-point of DC and FP || CQ.

By converse of mid-point theorem, we get :

P is the mid-point of DQ.

⇒ DP = PQ .....(1)

In △ APB,

E is the mid-point of AB and EQ || AP.

By converse of mid-point theorem, we get :

Q is the mid-point of BP.

⇒ PQ = QB ......(2)

From equations (1) and (2) we get :

⇒ PQ = QB = DP.

∴ AF and EC trisect BD.

Hence, proved that AF and EC trisect the diagonal BD.