In triangle $ABC$, the angle bisectors of angles $\alpha$ and $\gamma$ intersect at point $M$. Calculate angle $\beta$ if it is known that it is equal to half of the angle $AMC$.

1. Problem Description

In triangle $ABC$, the angle bisectors of angles $\alpha$ and $\gamma$ intersect at point $M$. Calculate angle $\beta$ if it is known that it is equal to half of the angle $AMC$.

2. Solution Steps

Let $\alpha$, $\beta$, and $\gamma$ be the angles of triangle $ABC$. The sum of the angles in a triangle is 180 degrees, so we have:

$$\alpha + \beta + \gamma = 180^{\circ}$$

Since $AM$ and $CM$ are angle bisectors of angles $\alpha$ and $\gamma$, we have $\angle MAC = \alpha/2$ and $\angle MCA = \gamma/2$.

In triangle $AMC$, the sum of the angles is 180 degrees, so we have:

$$\angle MAC + \angle AMC + \angle MCA = 180^{\circ}$$

$$\frac{\alpha}{2} + \angle AMC + \frac{\gamma}{2} = 180^{\circ}$$

$$\angle AMC = 180^{\circ} - \frac{\alpha}{2} - \frac{\gamma}{2}$$

$$\angle AMC = 180^{\circ} - \frac{\alpha + \gamma}{2}$$

We are given that $\beta = \frac{1}{2} \angle AMC$, so

$$\beta = \frac{1}{2} (180^{\circ} - \frac{\alpha + \gamma}{2})$$

$$\beta = 90^{\circ} - \frac{\alpha + \gamma}{4}$$

We know that $\alpha + \gamma = 180^{\circ} - \beta$. Substituting this into the equation:

$$\beta = 90^{\circ} - \frac{180^{\circ} - \beta}{4}$$

$$4\beta = 360^{\circ} - (180^{\circ} - \beta)$$

$$4\beta = 360^{\circ} - 180^{\circ} + \beta$$

$$4\beta = 180^{\circ} + \beta$$

$$3\beta = 180^{\circ}$$

$$\beta = \frac{180^{\circ}}{3}$$

$$\beta = 60^{\circ}$$

3. Final Answer

$\beta = 60^{\circ}$