The parallel sides of a trapezium are a and b respectively. The line joining the midpoint of its non parallel sides will be
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ANSWER
Let ABCD is a trapezium in which AB∥DC and E,F are mid-points of AD, BC respectively.
Join CE and produce it to meet BA produced at G.
It is given that p and q are parallel sides.
∴ AB=p and CD=q ----- ( 1 )
In △EDC and △EAG,
⇒ E=EA [ E is mid point of AD ]
⇒ ∠CED=∠GEC [ Vertically opposite angles ]
⇒ ∠ECD=∠EGA [ Alternate angles]
⇒ △EDC≅△EAG [ By ASA rule ]
⇒ CD=GA and EC=EG [ By C,P.C.T ]
In △CGB,
E is mid-point of CG [ Since, EC=EG ]
F is a mid-pont of BC
∴ By mid-point theorem EF∥AB and EF=
2
1
GB
But GB=GA+AB=CD+AB
Hence, EF∥AB and EF=
2
1
(AB+CD)
∴ EF=
2
1
(p+q) [ From ( 1 ) ]
∴ We have proved that, "the line joining the mid point of its non-parallel sides will be
2
p+q
.
Answer:
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