Trapezium Basics

What is a Trapezium?

Trapezium is a quadrilateral with one pair of parallel sides. Geometry

In the above figure, AB ∥ CD

Isosceles trapezium - A trapezium whose non-parallel sides are equal.
That is, AB ∥ CD and BC = AD

A parallelogram is a trapezium. However, vice-versa need not be true, i.e. a trapezium need not be a parallelogram - as only one pair of opposite sides is parallel in a trapezium and we require both pairs to be parallel in a parallelogram.




Properties of Trapezium

These are the properties that are true for any kind of trapezium.

Property 1: Angles

Consecutive angles along both non-parallel sides are supplementary, i.e. their sum is equal to 180°. Geometry

That is, if AB ∥ CD then:
∠A + ∠D = 180°, and
∠B + ∠C = 180°

We studied in Lines and Angles module, that:
Sum of interior angles (or exterior angles) on the same side of the transversal line is supplementary. Geometry

In case of Isosceles trapeziums, consecutive angles along each parallel sides are equal too. Geometry

That is, if AB ∥ CD and BC = AD, then:
∠A = ∠B, and
∠C = ∠D

Property 2: Diagonals

Diagonals of a trapezium intersect each other proportionally. Geometry

$\frac{AO}{OC}$ = $\frac{BO}{OD}$

In other words, if diagonals of a quadrilateral divide each other proportionally, then it must be a trapezium.

In case of Isosceles trapeziums, diagonals are equal too. Geometry

That is, if AB ∥ CD and BC = AD, then:
AC = BD

Property 3: A line parallel to the parallel sides

Any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally. Geometry

In the above figure, if AB ∥ PQ ∥ DC then,
$\frac{AP}{PD}$ = $\frac{BQ}{QC}$

It is an application of Thales theorem (or Basic proportionality theorem) that we read in triangle module.

Mid-point Theorem: A Special Case

If the line parallel to the parallel sides of a trapezium, passes through the mid points of non-parallel sides, then its length will be equal to half of the sum of parallel sides. Geometry

In the above figure, if $\frac{AP}{PD}$ = $\frac{BQ}{QC}$ = 1, i.e. AP = PD and BQ = QC, then:
PQ = $\frac{AB + DC}{2}$

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