Answer
$\angle ABC = 35^{\circ}$
$\angle ADC = 35^{\circ}$
$\angle 1 = 110^{\circ}$
Work Step by Step
By definition of isosceles triangles, the two angles $\angle ABD$ and $\angle ADB$ are equal. Also, the sum of the three angles of a triangle is $180^{\circ}$
Let $x$ be the measure of each angle $\angle ABD$ and $\angle ADB$
$x + x + 40^{\circ} = 180^{\circ}$
$2x = 180^{\circ}- 40^{\circ}$
$2x = 140^{\circ}$
$x = \frac{140^{\circ}}{2}$
$x = 70^{\circ}$
Then both $\angle ABD$ and $\angle ADB$ have a measure of $70^{\circ}$
Since $\overline{BC}$ bisects $\angle ABD$, we can find the measure of angle $\angle ABC$:
$\angle ABC = \frac{\angle ABD}{2} = \frac{70^{\circ}}{2} = 35^{\circ}$
Since $\overline{DC}$ bisects $\angle ADB$, we can find the measure of angle $\angle ADC$:
$\angle ADC = \frac{\angle ADB}{2} = \frac{70^{\circ}}{2} = 35^{\circ}$
The sum of $\angle CBD$, $\angle CDB$, and $\angle 1$ is $180^{\circ}$. We can find the measure of $\angle 1$:
$\angle CBD + \angle CDB + \angle 1 = 180^{\circ}$
$35^{\circ} + 35^{\circ} + \angle 1 = 180^{\circ}$
$70^{\circ} + \angle 1 = 180^{\circ}$
$\angle 1 = 180^{\circ} - 70^{\circ}$
$\angle 1 = 110^{\circ}$
