1. Prove that the line joining the mid-points of the diagonals of a trapezium is parallel
to the parallel sides of the trapezium.
2. P is the mid-point of the side CD of a parallelogram ABCD. A line through C
Oparallel to PA intersects AB at Q and DA produced at R. Prove that DA = AR and
CQ = QR.
Answers
Answer:
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Step-by-step explanation:
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 ) ]
solution
Answered By
toppr
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