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The answer is C however if angle ACD is 110 degrees and angle AB is 110 degrees how does it equal 180?

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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 $$\angle ACD = 110^\circ$$

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  • $\begingroup$ Where does the 2 come from? $\endgroup$ – Jon Feb 15 '13 at 19:51
  • $\begingroup$ Because the $\angle C = \angle B$, so $\angle A + \angle B + \angle C = \angle A + \angle C +\angle C = \angle A + 2\angle C = 180^\circ$ $\angle A + 2 \angle C = 180 \implies 40 + 2\angle C = 180 \implies 2\angle C = 180 - 40 = 140$ $\endgroup$ – Namaste Feb 15 '13 at 19:55
  • $\begingroup$ $\angle B = \angle C = 70^\circ$, and we have that the sum of angle A + angle B = $40 + 70 = 110 = \angle ACD$ $\endgroup$ – Namaste Feb 15 '13 at 20:00
  • $\begingroup$ the measure of an exterior angle is equal to the sum of the measures of interior angles opposing it. $\endgroup$ – Namaste Feb 15 '13 at 20:03
  • $\begingroup$ Is this clear now, Jon? $\endgroup$ – Namaste Feb 15 '13 at 20:11
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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 $\angle ACD = 110^{\circ}$.

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