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I am asked to get the value of $\theta$ in terms of $\alpha$ and $\beta$. Quick geogebra "shows" it is the difference, but how do you show it mathematically? All the properties I have about it are that the incenter is the intersection of the bisectors, and the orthocenter of the altitudes.

After putting some variables on the angles BAI and ABI, I expected those variables to get cancelated, but they did not. How can I poceed further?

chubakueno
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Express angles $\alpha$, $\beta$, $\theta$ in terms of angles $A$, $B$, $C$, observing that $$\angle BAI = \frac{1}{2}A \qquad \text{and} \qquad \angle BAO = 90^\circ - B$$

Orthocenter and Incenter

Because of how $AI$ and $AO$ are arranged about $A$, we have $$\alpha = (90^\circ - B) - \frac{1}{2}A = \frac{1}{2}\left( 180^\circ - 2 B - A\right) = \frac{1}{2}\left( C - B \right)$$

Find similar expressions for $\beta$ (in terms of $A$ and $C$) and $\theta$ (in terms of $A$ and $B$), noting that the arrangement of altitude and angle bisector at each vertex doesn't always match. (One will differ from the others (why?), and that's the key point!) Then see if, as you suspect, one of $\alpha$, $\beta$, $\theta$ is the difference of the other two.

Blue
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  • The suggested solution is really good. To make things a bit easier, I will re-write α as (1/2)[180 - B - A - B] and hence, α = (1/2) [C - A]. – Mick Aug 30 '13 at 16:38
  • α = (1/2) [C - B], my mistake. – Mick Aug 30 '13 at 17:03
  • @Mick: Good call. (I'm embarrassed to have overlooked it.) I added your simplification to my equation, and adjusted the following paragraph appropriately. – Blue Aug 30 '13 at 18:22
  • Thank you, I don´t know why I didn´t see this. Anyways, I think that with a little more thought, my initial approach should have worked, but indeed it would have been to messy to continue it :) – chubakueno Aug 30 '13 at 21:32