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 their difference ..

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Answer 1

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Let E and F are midpoints of the diagonals AC and BD of trapezium ABCD respectively.

Draw DE and produce it to meet AB at G

Consider △AEG and △CED⇒ ∠AEG=∠CED [ Vertically opposite angles ]

⇒ AE=EC [ E is midpoint of AC ]

⇒ ∠ECD=∠EAG [ Alternate angles ]

⇒ △AEG≅△CED [ By SAA congruence rule ]

⇒ DE=EG ---- ( 1 ) [ CPCT ]

⇒ AG=CD ----- ( 2 )

In △DGB

E is the midpoint of DG [ From ( 1 ) ]

F is midpoint of BD

∴ EF∥GB

⇒ EF∥AB [ Since GB is part of AB ]

⇒ EF is parallel to AB and CD.

Also, EF= 1/2 GB

⇒ EF= 1/2 (AB−AG)

⇒ EF= 1/2 (AB−CD) [ From ( 2 ) ]