**Angle Sum Property of a Triangle**

Triangle is the smallest polygon which has three sides and three interior angles.

In the given triangle, âˆ†ABC, AB, BC, and CA represent three sides. A, B and C are the three vertices and âˆ ABC, âˆ BCA and âˆ CAB are three interior angles of âˆ†ABC.

**Figure 1 Triangle ABC**

**Theorem 1:** Angle sum property of triangle states that the sum of interior angles of a triangle is 180Â°.

**Proof:Â **Consider a âˆ†ABC, as shown in the figure below. To prove the above property of triangles, draw a line \( \overleftrightarrow {PQ} \) parallel to the side BC of the given triangle.

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

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

SincePQ||BC and AB, AC are transversals,

Therefore, âˆ QAC = âˆ ACB (a pair of alternate angle)

Also, âˆ PAB = âˆ CBA (a pair of alternate angle)

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

âˆ ACB + âˆ BAC + âˆ CBA= 180Â°

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

**Exterior Angle Property of a Triangle:**

Theorem 2: If any side of a triangle is extended, then the exterior angle so formed is the sum of the two opposite interior angles of the triangle.

In the given figure, the side BC of âˆ†ABC is extended. The exterior angle âˆ ACD so formed is the sum of measures of âˆ ABC and âˆ CAB.

Proof: From figure 3, âˆ ACB and âˆ ACD forms a linear pair since they represent the adjacent angles on a straight line.

Thus, âˆ ACB + âˆ ACD = 180Â° Â ……….(2)

Also, from the angle sum property it follows that:

âˆ ACB + âˆ BAC + âˆ CBA = 180Â° ……….(3)

From equation (2) and (3) it follows that:

âˆ ACD = âˆ BAC + âˆ CBA

This property can also be proved using concept of parallel lines as follows:

In the given figure, side BCof âˆ†ABC is extended. A line \( \overleftrightarrow {CE} \)Â parallel to the side AB is drawn, then: Since \( \overline {BA} ~||~\overline{CE}\) and \( \overline{AC}\) is the transversal,

âˆ CAB = âˆ ACE Â ………(4) (Pair of alternate angles)

Also,Â \( \overline {BA} ~||~\overline{CE}\) and \( \overline{BD}\)iis the transversal

Therefore, âˆ ABC = âˆ ECD Â ……….(5) (Corresponding angles)

We have, âˆ ACB + âˆ BAC + âˆ CBA = 180Â° ………(6)

Since,the sum of angles on a straight line is 180Â°

Therefore, âˆ ACB + âˆ ACE + âˆ ECD = 180Â° ………(7)

Since, âˆ ACE + âˆ ECD = âˆ ACD(From figure 4)

Substituting this value in equation (7);

âˆ ACB + âˆ ACD = 180Â° ………(8)

From the equations (6) and (8) it follows that,

âˆ ACD = âˆ BAC + âˆ CBA

Hence it can be seen that the exterior angle of a triangle is equals to the sum of its opposite interior angles.

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