**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) Ã— 180Â°. Thus, the sum of interior angles of a triangle is 180Â° and the sum of interior angles of a quadrilateral is 360Â°. Let us try to prove this mathematically.

**Angle Sum Property of Triangles:**

Consider a âˆ†ABC, as shown in the figure below.

Construction: Draw a line \( \overleftrightarrow {PQ} \)

Figure 1: Proof of angle sum property of triangles

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

âˆ PAB + âˆ BAC + âˆ QAC = 180Â° ….(1)

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

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

Substituting the value of âˆ QAC andâˆ PAB in equation (1),

âˆ ACB + âˆ BAC + âˆ CBA = 180Â°

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

**Angle Sum Property of Quadrilaterals:**

Consider a quadrilateral ABCD

Constructions: Draw diagonal BD as shown below:

Figure 3

From figure 3;

âˆ B = âˆ 3 + âˆ 4 …..(1)

âˆ D = âˆ 1 + âˆ 2 …..(2)

In âˆ†ABD,

âˆ A + âˆ 1 + âˆ 3 = 180Â° â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(3) (Angle Sum Property of triangles)

Similarly, in âˆ†BCD,

âˆ C + âˆ 2 + âˆ 4 = 180Â°â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(4) (Angle Sum Property of triangles)

Adding equation (3) and (4);

â‡’ âˆ A + âˆ 1 + âˆ 3 + âˆ C + âˆ 2 + âˆ 4 = 360Â°

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

âˆ A + âˆ B + âˆ C + âˆ D = 360Â°

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

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

Also, quadrilateralABCD is divided into two triangles by diagonalBD, âˆ†ABD and âˆ†BCD. The measure of sum of interior angles of ABCD is twice the sum of interior angles of a triangle i.e. 2 Ã— 180Â°.’