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Question
ABCD is a trapezium in which side AB is parallel to side DC. P is the mid-point of side AD. IF Q is a point on the Side BC such that the segment PQ is parallel to DC, prove that PQ = `(1)/(2)("AB" + "DC")`.
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Solution
PQ || DC ⇒ OQ || DC ||AB
Therefore, Q and O are mid-points of BC and BD respectively.
In ΔABD,
P and O are mid-points of AD and BD respectively
⇒ OP = `(1)/(2)"AB"` ........(i)
In ΔBCD,
Q and O are mid-points of BC and BD respectively
⇒ OQ = `(1)/(2)"CD"` ........(ii)
Adding (i) and (ii)
OP + OQ = `(1)/(2)"AB" + (1)/(2)"CD"`
⇒ PQ = `(1)/(2)("AB" + "CD")`.
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