Question

prove that the line segment joining the midpoints of the diagonals of a trapezium is parallel to the parallel sides and equal to half the difference.​

Answer 1

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$\huge\underline\bold\orange{Question}$

prove that the line segment joining the midpoints of the diagonals of a trapezium is parallel to the parallel sides and equal to half the difference.

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$\huge\underline\bold\orange{Solution}$

Let ABCD is a trapezium in which AB || DC, and let M and N be the midpoints of the diagonals AC and BD respectively.

Join CN produce it to meet AB at E

In ∆CDN and EBN,we have;

• DN = BN [∴N is midpoint of BD]
• ∠DCN = ∠BEN (alt.interior ∠s)
• ∠CDN = ∠EBN (alt.interior ∠s)

∴∆CDN ≅ ∆EBN [SAA-criteria].

∴DC = EB and CN = NE (c.p.c.t.).

Thus, in ∆CAE,the points M and N are the midpoints of AC and DC respectively.

$\therefore \: MN || AE \: and \: MN = \frac{1}{2} AE = > MN || AB || DC$

$and \: MN \: = \frac{1}{2} AE = \frac{1}{2} (AB - EB) = \frac{1}{2} (Ab - DC) \: \: \: (\therefore \: EB = Dc)$

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$\sf{\underline{\underline{\pink{Additional\:Information-}}}}$

✯︎Trapezium a quadrilateral having one pair of opposite sides parallel is called

✯︎In trapezium KLMN , we have

• KL || NM.

✯︎The line segment joining the midpoints of non parallel sides of a trapezium is called its median

Answer 2

Answer:

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