Any closed figure in plane, bounded by four line segments is called a Quadrilateral. In the given quadrilateral $ABCD$: i) Four vertices are $A, \ B, \ C, \ and\ D$

ii) Four line segments are $AB,\ BC,\ CD,\ DA$

iii) Four angles are $\angle A,\ \angle B, \ \angle C \ and \ \angle D$

iv) Two diagonals are $AC \ and\ BD$

Let’s also learn the nomenclature related to quadrilaterals:

1) Adjacent Sides: Two sides of a quadrilateral having a common vertex. Example $AD \ and\ DC$ are adjacent sides with D as the common vertex. Similarly, $AB \ and\ BC$ are adjacent sides with common vertex B.

2) Opposite Sides: Two sides of a quadrilateral that doesn’t share a common vertex. Example: $AB \ and\ DC$ are opposite side. Similarly, $AD \ and\ BC$ are opposite sides. You will also see that they don’t share any common vertex.

3) Adjacent Angles: Two angles of a quadrilateral having a common arm are called adjacent angles Example: In the above figure $\angle A \ and\ \angle B$ are adjacent and they share $AB$ as the common arm. Similarly, $\angle B \ and\ \angle C$ are adjacent angles and they share $BC$ as the common arm.

4) Opposite Angles: Angles in a quadrilateral that doesn’t share any common arm are called opposite angles. Example: In the above figure, $\angle A \ and\ \angle C$ are opposite angles. Similarly, $\angle D \ and\ \angle B$ are opposite angles. $\\$

Important Theorem:

Sum of the interior angles $= (n-2) \times 180=(2n-4) \times 90^{\circ}$

In a quadrilateral, $n=4$

Substituting in the above formula, we get that the sum of the internal angles is $360^{\circ}$

Another way of proving this is as follows: In $\Delta ABC \colon \angle 2 + \angle 4 + \angle B = 180^{\circ} ...i)$

Similarly, in $\Delta ADC: \angle 1 + \angle 3 + \angle D = 180^{\circ} ...ii)$

Adding i) and ii) we get $(\angle 1 + \angle 2) + (\angle 4 + \angle 3) + \angle D + \angle B = 360^{\circ}$

Or $\angle A+\angle B+\angle C+\angle D=360^{\circ}$