# Finding angles from some other angles related to incircle

Let $ABC$ be a triangle and $O$ the center of its enscribed circle.

Let $M = BO \cap AC$ and $N=CO \cap AB$ such that $\measuredangle NMB = 30°, \measuredangle MNC = 50°$.

Find $\angle ABC, \angle BCA$ and $\angle CAB$.

I also posted this here at the Art of Problem Solving.

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At first, I took $O$ to denote the center of the incircle inscribed into the triangle. Only a comment below clarified that you were actually taling about the center of the circumcircle circumscribed around the triangle. Therefore, I have two solutions, one for each interpretation.

# Circumcircle

## Angles without a proof

I constructed your situation using Cinderella. At first I had one point chosen freely on a line, which I later moved into a position where everything fit the way it should. From the resulting figure, I obtained the following measurements:

\begin{align*} \measuredangle ABC &= 60° \\ \measuredangle BCA &= 70° \\ \measuredangle CAB &= 50° \end{align*}

Warning: I accidentially swapped $M$ and $N$ as well as $B$ and $C$ in the following image. So take care to look more at the actual angles than the denoted point names.

I guess that once you know these angles at the corners, you can choose suitable coordinates and using these proove that the angles specified in your question are indeed as required.

## Oriented angles

In the above, I interpreted the angles in your question as unoriented, and measured one of the clockwise and the other counter-clockwise. I furthermore assumed $M$ and $N$ to lie between the corresponding corners of the triangle. Strictly speaking, your angles are given in an oriented fashion, and as both of them are positive, $B$ and $C$ must lie on different sides of $MN$. this leads to rather ugly triangles. Of the two possible solutions I found, the better one is the following:

The angles for this triangle are rather ugly, compared to the nice numbers resulting from the interpretation above.

# Incircle

## Angles without a proof

This result I obtained in a similar way to the one outlined above, using Cinderella with one free point on a line adjusted till things line up as intended.

\begin{align*} \measuredangle ABC &= 40° \\ \measuredangle BCA &= 120° \\ \measuredangle CAB &= 20° \end{align*}

## Construction

Let us call the half corner angles like this:

\begin{align*} \alpha &= \angle BAO = \angle OAC \\ \beta &= \angle CBO = \angle OBA \\ \gamma &= \angle ACO = \angle OCB \end{align*}

From $\triangle BCO$ you see that $\angle BCO = 180° - \beta - \gamma$. That is the same as $\angle MON$, so from $\triangle MNO$ you can conclude that $\beta+\gamma = 30° + 50°$. From $\triangle ABC$ you know that $2\alpha + 2\beta + 2\gamma = 180°$, so $\alpha = 90° - (\beta + \gamma) = 10°$.

Based on this angle, you can construct the triangle even without knowing the corner angles up front:

1. Choose $M$ and $N$ arbitrarily
2. Draw the rays $MO$ and $NO$ using the given angles to obtain $O$
3. Draw lines at $90° - \alpha = 80°$ from $MO$ through $M$ and $O$ as indicated in the figure below (green). Around their intersection, draw a circle containing all points that see $M$ and $O$ under an angle of $10°$.
4. Repeat the previous step for $NO$. The intersection of these two circles is $A$.
5. Now $C = AM \cap NO$ and $B = AN \cap MO$.

Measuring the angles in this figure will give you the desired result up to the precision with which you performed your drawings and measurements.

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I'm sorry, I think I used the wrong notion. I mean that $O$ is the circumcenter, the center of the circle passing through the three vertices of the triangle – Andrei Oct 22 '12 at 12:30
@Andrei: If you're talking about circumcircle instead of incircle, that changes the problem fundamentally. I didn't know the word “enscribed” in this context, and so assumed “inscribed”. Will think about the other task. – MvG Oct 22 '12 at 14:47
@Andrei, please have a look at my revised answer. – MvG Oct 23 '12 at 12:04
I guess the values are corect, but can we find them without the software? – Andrei Oct 30 '12 at 7:57