Question

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

Answer 1

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Answer 2

Correct Question:

ABCD is a parallelogram. E and F are the mid -points of the sides AB and CD respectively. Prove that the line segments AF and CE trisect the diagonal BD.

Answer:

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

$\therefore$ $\sf{AF=}$$\sf\dfrac{1}{2}AB\:and\:CF=$$\sf\dfrac{1}{2}CD$ ___________(i)

But, ABCD is a llgm

$\therefore$ $\tt\underline{AB=CD\:and\:AB||DC}$

$\Rightarrow$ $\sf\dfrac{1}{2}AB=$$\sf\dfrac{1}{2}CD\:and\:AB||DC$

$\Rightarrow$ $\tt{AE=FC\:and\:AE||FC}$________ (from (i))

$\Rightarrow$ $\tt{AECF\:is\:a\:llgm}$

$\Rightarrow$ $\tt{FA=CE}$

$\Rightarrow$ $\tt{FQ||CP}$______(ii)

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

____________________________________

⌣ In ∆DCP, F is the mid-point of CD and FQ || CP.

$\therefore$ $\sf{Q\:is\:the\:mid-point\:of\:DP}$ _____________(from (ii))

$\implies$ $\tt\underline{PQ=DP}$ ___________(iii)

___________________________________________________

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

$\implies$ $\sf{P\:is\:the\:mid-point\:of\;BQ}$

$\implies$ $\tt\underline{BP=PQ}$ _____(iv)

____________________________________

◕ From (iii) and (iv), we get

$\qquad\sf\underline{BP=PQ=QD}$

$\implies$ P and q trisect BD.

$\implies$ AF and CE trisect BD.

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