Question

prove that the straight line joining the midpoints of the diagonals of a trapezium is parallel to the parallel sides of the trapezium and is equal to half of their difference

Answer 1

Let ABCD is a trapezium in which AB || CD.

Let P and Q are the mid points of the diagonals AC and BD respectively.

We have to prove that:

PQ || AB or CD and

PQ = (AB - CD)/2

Since AB || CD and AC cuts them at A and C, then

∠1 = ∠2 (alternate angles)

Again from ΔAPR and ΔDPC,

∠1 = ∠2 (alternate angles)

AP = CP (since P is the mid=point of AC)

∠3 = ∠4 (vertically opposite angles)

From ASA congruent rule,

ΔAPR ≅ ΔDPC

Then from CPCT,

AR = CD and PR = DP

Again in ΔDRB, P and Q are the mid points of the sides DR and DB,

then PQ || RB

=> PQ || AB

=> PQ || AB and CD

Again in ΔDRB, P and Q are the mid points of the sides DR and RB,

then PQ = RB/2

=> PQ = (AB - AR)/2

=> PQ = (AB - CD)/2

Answer 2

Step-by-step explanation:

In trapezium ,ABCD,AB Parallel to CD and p,q are mid points of AC and BD

we nwe'd to prove pq parallel to AB and DC

pq=1/2(AB-DC)

construction:join DP and produce DP to meet AB in R