Question

prove exterior angle property of triangle ​

Answer 1

Answer:

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 ∆ABC is extended. The exterior angle ∠ACD so formed is the sum of measures of ∠ABC and ∠CAB.

Answer 2

Answer:

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 ∆ABC is extended. The exterior angle ∠ACD so formed is the sum of measures of ∠ABC and ∠CAB.

Proof:

From figure 3, ∠ACB and ∠ACD form a linear pair since they represent the adjacent angles on a straight line.

Thus, ∠ACB + ∠ACD = 180°  ……….(2)

Also, from the angle sum property, it follows that:

∠ACB + ∠BAC + ∠CBA = 180° ……….(3)

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

∠ACD = ∠BAC + ∠CBA

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

In the given figure, side BC of ∆ABC is extended. A line CE←→ parallel to the side AB is drawn, then: Since BA¯¯¯¯¯¯¯¯ || CE¯¯¯¯¯¯¯¯ and AC¯¯¯¯¯¯¯¯ is the transversal,

∠CAB = ∠ACE   ………(4) (Pair of alternate angles)

Also, BA¯¯¯¯¯¯¯¯ || CE¯¯¯¯¯¯¯¯ and BD¯¯¯¯¯¯¯¯ is the transversal

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

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

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

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

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

Substituting this value in equation (7);

∠ACB + ∠ACD = 180° ………(8)

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

∠ACD = ∠BAC + ∠CBA

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