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Question
The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that:
PR = `[1]/[2]` ( AB + CD)
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Solution
Here from the triangle,
ABD P is the midpoint of AD and PR || AB,
therefore Q is the midpoint of BD
Similarly, R is the midpoint of BC as PR || CD || AB
From triangle ABD,
PQ = `1/2` AB ...(1) ...[by Mid-point theorem]
From triangle BCD,
QR = `1/2` CD ...(2) ...[by Mid-point theorem]
Now (1) + (2)
PQ + QR = `1/2 "AB" + 1/2 "CD"`
PR = `1/2`(AB + CD)
Hence proved.
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