Frustrating problem

• Jun 29th 2007, 07:09 PM
pianistonstrike
Frustrating problem
Sorry if this seems really simple... I haven't had geometry in school in over a year.

The bisectors of angles B and C in triangle ABC intersect in point O. Find angle BOC if angle BAC = a (pretend that's an alpha).

Bleh, I translated this from Russian, sorry if anything's unclear. Any help is appreciated though!
• Jun 29th 2007, 07:39 PM
Jhevon
Quote:

Originally Posted by pianistonstrike
Sorry if this seems really simple... I haven't had geometry in school in over a year.

The bisectors of angles B and C in triangle ABC intersect in point O. Find angle BOC if angle BAC = a (pretend that's an alpha).

Bleh, I translated this from Russian, sorry if anything's unclear. Any help is appreciated though!

See the diagram below.

I will not use three letters to describe an angle, i'll use one.

Let O be angle BOC
Let $\displaystyle \alpha$ be angle BAC
Let C be angle ACB
Let B be angle ABC

We want to find angle O (the red angle) in terms of $\displaystyle \alpha$

Remember that all angles in a triangle add up to $\displaystyle 180^{ \circ}$, and that bisecting an angle means dividing it into two equal parts.

So, for $\displaystyle \triangle ABC$ we must have that:

$\displaystyle \alpha + B + C = 180$

$\displaystyle \Rightarrow B + C = 180 - \alpha$ .................(1)

Now for $\displaystyle \triangle BOC$ we must have:

$\displaystyle O + \frac {1}{2}B + \frac {1}{2}C = 180$

$\displaystyle \Rightarrow O = 180 - \frac {1}{2}B - \frac {1}{2}C$

$\displaystyle \Rightarrow O = 180 - \frac {1}{2}(B + C)$

Now substitute the expression for B + C from equation (1), we get:

$\displaystyle O = 180 - \frac {1}{2} (180 - \alpha )$

$\displaystyle \Rightarrow O = 180 - 90 + \frac {1}{2} \alpha$

$\displaystyle \Rightarrow \boxed { O = 90 + \frac {1}{2} \alpha }$
• Jun 29th 2007, 07:49 PM
Krizalid
Remember property of every concave quadrilateral and we have that (in Jhevon's sketch, where I denote $\displaystyle \measuredangle~A=2\alpha$)

$\displaystyle \measuredangle~\theta=2\alpha+\beta+\gamma=90^\cir c+\frac{\measuredangle~A}2~\blacksquare$
• Jun 29th 2007, 07:54 PM
pianistonstrike
Oh... duh, that makes perfect sense! I had gotten several of those steps but couldn't figure out how to tie them together.

Thank you so much!