Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to each of the parallel sides and is equal to half the difference of these sides.
Since, AB || DC and transversal AC cuts then at A and C respectively.
∴∠1 = ∠2 (Alternate angles)
Now, In ΔAPR and ΔDPC,
∠1 = ∠2
AP = CP (P is mid point of AC)
∠3 = ∠4 (Vertically opposite angles)
∴ ΔAPR ΔDPC
⇒ AR = DC and PR = DP
In ΔDRB, P and Q are the mid-points of sides DR and DB respectively.
∴ PQ || RB
⇒ PQ || AB
⇒ PQ || AB and DC
Again P and Q are the mid-points of sides DR and DB respectively.