Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to each of the - Maths - Quadrilaterals

Question

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

Santosh Kumar · Accepted Answer

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.

$∴ PQ = \frac{1}{2} R B \Rightarrow PQ = \frac{1}{2} (AB - AR) \Rightarrow PQ= \frac{1}{2} (AB - DC)$

171

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.