Question

prove that line segment joining the midpoints of the diagonals of a trapezium is :-. a) parallel to the other to parallel diagonals. b) is equal to half of the difference of the parallel two sides take the midpoints as E,F. PLS HELP ME I NEED IT FOR EXAMS ​

Answer 1

i hope it helps you ☺️..

thank you. .

Answer 2

Step-by-step explanation:

ANSWER

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=

2

1

GB

⇒ EF=

2

1

(AB−AG)

⇒ EF=

2

1

(AB−CD) [ From ( 2 ) ]