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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.


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Solution

Step 1: Find the relation between DP and DQ.

Given: In ABCD, E is the mid-point of AB. F is the mid-point of CD.

ABCDAB2CD2AEFCAB=CDAB2=CD2AE=FC

Since two sides are equal and parallel AEFC is a parallelogram.

In DQC and DPF,

D=D [Common angle]

DPF=DQC [Alternate angles]

DFP=DCQ [Alternate angles]

By AAA similarity, DQC~DPF

DPDQ=DFCD

DPDQ=12 [F is the midpoint of CD]

P is the midpoint of DQ.

From the figure,

DP=PQ………………..1

Step 2: Find the relation between BP and BQ.

In ABP and QBE,

B=B [Common angle]

BQE=BPA [Alternate angles]

QEB=PAB [Alternate angles]

By AAA similarity, ABP~QBE

BQBP=BEAB

BQBP=12 [E is the midpoint of CD]

Q is the midpoint of BP.

From the figure,

PQ=BQ………………..2

Step 3: Prove the given condition.

From equation 1 and 2,

DP=PQ=BQ

Therefore, from the figure, we can say that, AF and EC trisect the diagonal BD.

Hence proved that the line segments AF and EC trisect the diagonal BD.


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