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 their difference

Here is the link for answer to your query.

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- 6

**Join DE and produce to meet AB at G **

**In triangle AEG and CEG**

**<1 = <2 (Alternate interior angles)**

**AE = EC (E is the mid point of AC)**

**<4 = <3 ( Vertically opposite angles)**

**Triangle AEG is congruent to triangle CED (by ASA congruence rule) **

**DE = EG ( CPCT)**

**AG = CD ( CPCT)**

**In triangle BGD**

**E is the mid point of DG**

**F is the mid point of BD**

**By mid point theorem, **

**EF II BG**

**EF II AB**

**AB II CD **

**Hence, AB II EF II CD**

**EF = 1/2 BG**

**EF = 1/2 ( AB - AG)**

**EF = 1/2 (AB - CD)**

**HOPE THIS HELPS :)**

- 30

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- -1

ABCD is a trapezium with AB parallel to CD and midpoints of AC and BD be P and Q

repectively

CONSTRUCTION - Draw a line from D passing through Pand meeting AB in E.

In triangle APE and triangle CPD

angle PAE = angle PCD ( alternate angles )

Angle PEA = angle PDC ( alternate angles )

AP = PC ( P is the midpoint of AC )

triangle APE is congruent to triangle CPD

PE = PD ( CPCT )

In triangle BED

P is the midpoint of DE ( proved above )

Q is the midpoint of BD ( given )

Therefore PQ is parallel to BE ( midpoint theorem )

SO PQ is parallel to AB

AB parralel to CD ( given )

SO PQ is parallel to CD and AB

HENCE PROVED

- 5