**Quadrilaterals and Triangles: Angle sum property**

A plane figure bounded by finite line segments to form a closed figure is defined as polygon.Based on the number of sides or vertices a polygon can be classified as n-sided polygon. A polygon bounded by three finite line segments is known as triangle. It is the smallest possible polygon. A four-sided polygon is known as a quadrilateral.

The sum of interior angles of any n-sided polygon is given by (n-2) × 180°. Thus, the sum of interior angles of a triangle is 180° and the sum of interior angles of a quadrilateral is 360°. Let us try to prove this mathematically.

**Angle Sum Property of Triangles:**

Consider a ∆ABC, as shown in the figure below.

Construction: Draw a line \( \overleftrightarrow {PQ} \) passing through point A such that PQ||BC .

Figure 1: Proof of angle sum property of triangles

Since PQ is a straight line, it can be concluded that:

∠PAB + ∠BAC + ∠QAC = 180° ….(1)

Since PQ||BC and AB, AC are transversals to the parallel lines,

Also, ∠QAC = ∠ACB and ∠PAB = ∠CBA (pair of alternate angles)

Substituting the value of ∠QAC and∠PAB in equation (1),

∠ACB + ∠BAC + ∠CBA = 180°

Thus, the sum of interior angles of a triangle is always 180°.

**Angle Sum Property of Quadrilaterals:**

Consider a quadrilateral ABCD

Constructions: Draw diagonal BD as shown below:

Figure 3

From figure 3;

∠B = ∠3 + ∠4 …..(1)

∠D = ∠1 + ∠2 …..(2)

In ∆ABD,

∠A + ∠1 + ∠3 = 180° …………………………(3) (Angle Sum Property of triangles)

Similarly, in ∆BCD,

∠C + ∠2 + ∠4 = 180°………………………….(4) (Angle Sum Property of triangles)

Adding equation (3) and (4);

⇒ ∠A + ∠1 + ∠3 + ∠C + ∠2 + ∠4 = 360°

Substituting ∠B and ∠DFrom equation (1) and (2) respectively, we have:

∠A + ∠B + ∠C + ∠D = 360°

Thus, it can be seen that the sum of interior angles in a quadrilateral is always 360°.

According to exterior angle property of a polygon, sum of exterior angles of a polygon is always 360° and hencethe sum of exterior angles of atriangle and a quadrilateral is 360°.

Also, quadrilateralABCD is divided into two triangles by diagonalBD, ∆ABD and ∆BCD. The measure of sum of interior angles of ABCD is twice the sum of interior angles of a triangle i.e. 2 × 180°.