Question about Geometry involving angles and lines 
The answer is C however if angle ACD is 110 degrees and angle AB is 110 degrees how does it equal 180?
 A: Since it's an isosceles triangle, angles $B, C$ (the angles opposite segments $AC, AB$  respectively, which are equal in length) are equal in measure: 
$$ \angle B = \angle C$$
We also know the the sum of the measures of the angles of a triangle sum to $180^\circ$.
$$\angle A + \angle B + \angle C = \angle A + 2 \angle C = 180^\circ $$ $$\implies 40^\circ + 2 \angle C = 180^\circ \implies 2\angle C = 180^\circ - 40^\circ = 140^\circ$$
Solve for (the measure of) angle $C$: $$\angle B = \angle C = \dfrac{140^\circ}{2} = 70^\circ$$ 
Since $\angle C$ and $\angle ACD$ (the unknown angle) are supplementary angles (since $B, C, D$ are colinear: they lie on the same straight line), we know that the sum of the measures of $\angle C$ and $\angle ACD$ is $180^\circ$ 
$$\angle C + \angle ACD = 180^\circ\implies \angle ACD = 180^\circ - \angle C = 180^\circ - 70^\circ = 110^\circ$$
The answer is, indeed, that the measure of $$\angle ACD = 110^\circ$$
A: The internal angles of a triangle add up to $180^{\circ}$. 
Since $\angle BAC = 40^{\circ}$ it follows that $\angle ABC + \angle ACB = 140^{\circ}$. Moreover, since $AB=AC$ it follows, by symmetry, that $\angle ABC = \angle ACB$. Hence: $\angle ABC = \angle ACB = 70^{\circ}$.
Since $B$, $C$ and $D$ are colinear, it follows that $\angle ACB + \angle ACD = 180^{\circ}$. We know that $\angle ACB = 70^{\circ}$ and hence $\angle ACD = 110^{\circ}$.
