In triangle $ABC$, the angle bisector $CD$ of angle $\gamma$ intersects side $AB$ at an angle $\varphi = 110^{\circ}$. Determine the angles of the triangle, given that $CD = BC$.

1. Problem Description

In triangle $ABC$, the angle bisector $CD$ of angle $\gamma$ intersects side $AB$ at an angle $\varphi = 110^{\circ}$. Determine the angles of the triangle, given that $CD = BC$.

2. Solution Steps

Let $\alpha, \beta, \gamma$ be the angles at vertices $A, B, C$ respectively. Since $CD$ is the angle bisector of $\gamma$, we have $\angle ACD = \angle BCD = \frac{\gamma}{2}$.

Also, the angle between $CD$ and $AB$ is given as $\varphi = 110^{\circ}$. Therefore, $\angle CDA = 110^{\circ}$ or $\angle CDB = 110^{\circ}$. We consider the case when $\angle CDB = 110^{\circ}$.

Since the sum of angles in triangle $BCD$ is $180^{\circ}$, we have:

$\angle BCD + \angle CDB + \angle CBD = 180^{\circ}$

$\frac{\gamma}{2} + 110^{\circ} + \beta = 180^{\circ}$

$\beta + \frac{\gamma}{2} = 70^{\circ}$

Also, we are given that $CD = BC$, which means that triangle $BCD$ is an isosceles triangle. Since $CD = BC$, we have $\angle CDB = \angle CBD$, thus $\beta = \angle CDB$ is not possible. Therefore, $\angle CDB \ne 110^{\circ}$. This means $\angle CDA = 110^{\circ}$.

Then, $\angle CDB = 180^{\circ} - 110^{\circ} = 70^{\circ}$.

Since $BC = CD$, triangle $BCD$ is an isosceles triangle, so $\angle CBD = \angle CDB = \beta$. Then, $\beta = \angle BDC$.

Thus $\angle BCD + \angle CDB + \angle CBD = 180^{\circ}$

$\frac{\gamma}{2} + \beta + \beta = 180^{\circ}$

$\frac{\gamma}{2} + 2\beta = 180^{\circ}$

Since $\angle CDB = 70^{\circ}$, we have $\beta = \angle CBD = \angle CDB = 70^{\circ}$.

Substituting this into the equation above,

$\frac{\gamma}{2} + 2(70^{\circ}) = 180^{\circ}$

$\frac{\gamma}{2} + 140^{\circ} = 180^{\circ}$

$\frac{\gamma}{2} = 40^{\circ}$

$\gamma = 80^{\circ}$

We know that the sum of angles in triangle $ABC$ is $180^{\circ}$.

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

$\alpha + 70^{\circ} + 80^{\circ} = 180^{\circ}$

$\alpha = 180^{\circ} - 150^{\circ} = 30^{\circ}$

The angles of the triangle are $\alpha = 30^{\circ}$, $\beta = 70^{\circ}$, and $\gamma = 80^{\circ}$.

3. Final Answer

$\alpha = 30^{\circ}$, $\beta = 70^{\circ}$, $\gamma = 80^{\circ}$