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)

Answer 1

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.