Question

the sides BC,CA and BA of triangle ABC are produced in order forming exterior angles ACD,BAE and CBF. show that angles ACD+BAE+CBF=360

Answer 1

Measure of exterior angles equals to $180{}^{\circ}$ minus measure of inner angle. Thus

$$\angle BCD = 180{}^{\circ} - \angle BCA$$ $$\angle BAE = 180{}^{\circ} - \angle BAC$$ $$\angle CBF = 180{}^{\circ} - \angle CBA$$

Adding these 3 equalities:

$$\begin{array}{l} \angle BCD + \angle BAE + \angle CBF = (180{}^{\circ} - \angle BCA) + (180{}^{\circ} - \angle BAC) + (180{}^{\circ} - \angle CBA) \\ = 180{}^{\circ} + 180{}^{\circ} + 180{}^{\circ} - (\angle BCA + \angle BAC + \angle CBA) \\ = 540{}^{\circ} - (\angle BCA + \angle BAC + \angle CBA) \\ \end{array}$$

Since sum of all angles in the triangle equals to $180{}^{\circ}$, we have:

$$\angle BCA + \angle BAC + \angle CBA = 180{}^{\circ}$$

Thus

$$\angle BCD + \angle BAE + \angle CBF = 540{}^{\circ} - (\angle BCA + \angle BAC + \angle CBA) = 540{}^{\circ} - 180{}^{\circ} = 360{}^{\circ}$$

The proof is finished.