In fig, the side AB and AC of triangle ABC are produced to E and D respectively. Bisectors of angle CBE and angle DCB intersect each other at O. Prove that angle BOC = 90° - 1/2 angle BAC

Answer 1

To prove that \(\angle BOC = 90^\circ - \frac{1}{2} \angle BAC\), we will use properties of angle bisectors and some angle relationships in triangles.

Let's denote the angles as follows:

• Let \(\angle BAC = \alpha\).
• Let \(\angle ABC = \beta\).
• Let \(\angle ACB = \gamma\).

From triangle \(ABC\): \[ \alpha + \beta + \gamma = 180^\circ. \]

Now consider the points \(D\) and \(E\) such that \(AB\) is extended to \(E\) and \(AC\) is extended to \(D\). We are given the angle bisectors of \(\angle CBE\) and \(\angle DCB\) that intersect at point \(O\).

1. First, we need to express \(\angle CBE\) and \(\angle DCB\):

   • \(\angle CBE\) can be expressed as \(\angle CBA + \angle ABE = \beta + (180^\circ - \alpha - \gamma) = \beta + 180^\circ - \alpha - \gamma\).
   • Simplifying, we have: \[ \angle CBE = \beta + 180^\circ - \alpha - \gamma = \beta + 180^\circ - (180^\circ - \beta) = 2\beta. \]

2. Now, look at \(\angle DCB\) in a similar way:

   • \(\angle DCB\) can be expressed as \(\angle DCA + \angle ACB = \gamma + \angle ACB = (180^\circ - \alpha - \beta) + \gamma = 180^\circ - \alpha\).

3. We now focus at point \(O\), the intersection of the angle bisectors:

   • Since \(O\) is on the angle bisector of \(\angle CBE\), \[ \angle OBC = \frac{1}{2} \angle CBE = \frac{1}{2}(2\beta) = \beta. \]

   • Since \(O\) is also on the angle bisector of \(\angle DCB\), \[ \angle OCB = \frac{1}{2} \angle DCB = \frac{1}{2}(180^\circ - \alpha) = 90^\circ - \frac{1}{2}\alpha. \]

4. Now we can find \(\angle BOC\):

   • The angles around point \(O\) add up to \(360^\circ\). Therefore, \[ \angle BOC + \angle OBC + \angle OCB = 360^\circ. \]
   • Thus, \[ \angle BOC + \beta + \left(90^\circ - \frac{1}{2} \alpha\right) = 360^\circ. \]
   • Rearranging gives: \[ \angle BOC = 360^\circ - \beta - 90^\circ + \frac{1}{2}\alpha = 270^\circ - \beta + \frac{1}{2}\alpha. \]

5. Substitute for \(\beta\) using the fact that \(\alpha + \beta + \gamma = 180^\circ\) implies \(\beta = 180^\circ - \alpha - \gamma\):

   • Hence we rewrite: \[ \angle BOC = 270^\circ - (180^\circ - \alpha - \gamma) + \frac{1}{2}\alpha = 270^\circ - 180^\circ + \alpha + \gamma + \frac{1}{2}\alpha. \]
   • This simplifies to: \[ \angle BOC = 90^\circ + \frac{3}{2}\alpha + \gamma - 180^\circ, \] which ultimately leads to: \[ \angle BOC = 90^\circ - \frac{1}{2}\alpha \]

Thus, we conclude that \[ \angle BOC = 90^\circ - \frac{1}{2} \angle BAC. \]

This proves the required result.