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

PR = 1/2 (AB + CD) (Proved)

Step-by-step explanation:

See the attached diagram.

Join the diagonal of the trapezium AC and it meets PR at S.

Considering Δ ACD, P is the midpoint of AD and PS is parallel to CD.

So, S is the midpoint of AC and hence, PS = 1/2 CD

Again, considering Δ ABC, S is the midpoint of AC and SR is parallel to AB.

So, R is the midpoint of BC and hence, RS = 1/2 AB

Therefore, PR = PS + SR = 1/2 CD + 1/2 AB = 1/2 (AB + CD) (Proved)

Answer 2

*answer* -

In ∆ ABD And ∆BCD

PQ=1/2AB(mid point theorem)

OR =1/2DC(Mid point theorem)

since, PQ+QR=PR

Therefore,. 1/2AB +1/2DC=PR

PR=1/2(AB+DC)

Step-by-step explanation:

HOPE IT'S HELPFUL!!