Question

5. In a parallelogram ABCD. E and F are the
mid-points of sides AB and CD respectively
(see Fig. 8.31). Show that the line segments AF
and EC trisect the diagonal BD.​

Answer 1

hey mate

here is ur answer

Given: ABCD is a parallelogram

E is the mid point of AB

AE=EB

F is the mid point of CD

DF=FC

AD parallel CB, AD=CB , DC=AB,DC parallel AB

To Prove: DP=PQ=QB=1/3BD

Proof : DC=AB and DC||AB (given)

=> 1/2 DC=1/2AB

=> 1/2DC||1/2AB

AE=FC, FC||AE

=>AEFC Is a ||gram

(because pair if opposite side is equal And parallel)

=>AF||EBC

In triangle DQC

F is mid point of DC (Given)

AF||EC (Proved)

=>PF||QC

Converse of mid point theorem

DP=PQ (1)

In triangle ABP

E is the mid point of AB

AF||EC

=>AP||EQ

By converse of mid point theorem

=> PQ=QB (2)

USING 1 AND 2

DP=PQ=QB (3)

DP+PQ+QB=DB

USING 3

3DP=1/2DB

=>DP=PQ=QB=1/3BD

Proved

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