Question

In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.

Answer 1

given abcd is a parallelogram and cd is parallet to ab and cd = ab
to prove that  dp=pq=qb
fc=ae [f and e are midpoints of sides cd and ab]
fc//ab[cd//ab]
in triangle cdq  f is the midpoint of side cd and fp//cq
therefore p is the midpoint of side dq
ie dp=pq
similarly we can prove pq=qb
so dp=pq=qb
ie af and ec trisect bd


it is a sure qn.i had it for my xam

Answer 2

SOLUTION:

Since E and F are the mid-points of AB and CD respectively.

∴ AE = 1/2AB and CF = 1/2CD _____(i)

But, ABCD is a parallelogram.

∴ AB = CD and AB || CD.

→ 1/2AB = 1/2CD and AB || DC

→ AE = FC and AE || FC ______[From(i)]

→ AECF is a parallelogram.

→ FA || CE

→ FQ || CP ___________(ii)

We know that the segment drawn through the mid-point of one side of a triangle and parallel to other side bisects the third side.

In △DCP, F is the mid-point of CD and FQ || CP _______[From(ii)]

∴ Q is the mid-point of DP

→ PQ = QD ___________(iii)

Similarly, in △ABQ, E is the mid-point of AB and EP || AQ.

→ P is the mid-point of BQ.

→ BP = PQ ___________(iv)

From (iii) and (iv), we get

BP = PQ = QD

→ P and Q trisect BD

→ AF and CE trisect at BD.