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

Given:

A trapezium ABCD in which AB || DC and p and q are the midpoints of the diagonals AC and BD respectively.

To prove:

1) PQ || AB or DC

2) PQ = 1 ( AB - DC )

2

Proof :

Since, AB || DC and transversal CA cuts them at A and C respectively,

angle 1 = angle 2 ...(1) (alternative angles)

Now, in triangles APR and DPC,

angle 1 = angle 2 ...{ from (1) }

AP = CP ...( P is the midpoint of AC )

angle 3 = angle 4 ...( vertically opposite angles )

→ APR ~= DPC by ASA criterion

→ AR = DC and PR = DP ...( CPCT )

In DRB, P and Q are main points of sides DR and DB respectively respectively.

→ PQ || RB

→ PQ || AB

→ PQ || AB and DC

Again, P and Q are midpoints of DR and RB respectively in DRB.

→ PQ = 1 RB

2

→ PQ = 1 ( AB - AR )

2

→ PQ = 1 ( AB - DC )

2

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