Discussion
Sides AB and AC of a triangle ABC are produced to points D and E respectively. Bisectors of angles CBD and BCE intersect each other at the point O. If ∠ BAC = 50°, then ∠ BOC is equal to :
Answer: D
\(\angle BOC\) is 65°.
O is where the bisectors of the two exterior angles at B and C meet. With \(\angle B\) and \(\angle C\) the interior angles of the triangle:
\[ \begin{aligned} \angle OBC &= \tfrac{1}{2}(180^{\circ} - B) \\ \angle OCB &= \tfrac{1}{2}(180^{\circ} - C) \end{aligned} \]
\[ \begin{aligned} \angle BOC &= 180^{\circ} - \angle OBC - \angle OCB \\ &= \tfrac{1}{2}(B + C) \\ &= \tfrac{1}{2}(180^{\circ} - 50^{\circ}) = 65^{\circ} \end{aligned} \]
This is the standard result \(\angle BOC = 90^{\circ} - \frac{A}{2}\) for exterior bisectors. 115° is \(90^{\circ} + \frac{A}{2}\), the result for the bisectors of the interior angles, which is the usual mix-up.
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