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# Prove that segment joining the mid …

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Prove that segment joining the mid point of non parellel side of trapezium is parellel to the parellel side and it's length is equal to half of the sum of lengths of parellel sides

K C 1 month, 3 weeks ago

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=21​GB ⇒  EF=21​(AB−AG) ⇒  EF=21​(AB−CD)           [ From ( 2 ) ]

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