Question

When the side of a triangle is produced the exterior angle formed is equal to the sum of the opposite interior angles .prove the theorem.

Answer 1

ANSWER to Question #90733 – Math – Geometry

Let $\alpha, \beta$ and $\gamma$ be the interior angles of $\Delta ABC$ (as shown in the figure).

The sum of three interior angles of the triangle is

$$\alpha + \beta + \gamma = 180{}^\circ \text{...(1)}$$

Also, the line BC is produced to D hence BD is a straight line and ray CA stands on it.

Hence the sum of interior angle $\gamma$ and the exterior angle $\delta$ (as shown in the figure) is equal to $180{}^\circ$,

i.e.

$$\delta + \gamma = 180{}^\circ \text{...(2)}$$

From (1) and (2) it follows that

$$\delta + \gamma = \alpha + \beta + \gamma \Rightarrow \delta = \alpha + \beta$$

i.e. the formed exterior angle is equal to the sum of the opposite interior angles.

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