2007 AMC 10A Problem 8

Draw it first: $B$ at the top, $A$ and $C$ at the bottom, and $D$ inside the triangle, directly above the middle of $AC$. Both triangles stand on the same bottom side $AC$.

1. Base angle of the big triangle. In triangle $ABC$, the two equal sides meet at $B$ with angle $40^\circ$. The other two angles are equal and share what is left:
$$\angle BAC = \frac{180^\circ - 40^\circ}{2} = 70^\circ.$$
2. Base angle of the small triangle. In triangle $ADC$, the equal sides meet at $D$ with angle $140^\circ$:
$$\angle DAC = \frac{180^\circ - 140^\circ}{2} = 20^\circ.$$
3. Put them together at $A$. Since $D$ is inside triangle $ABC$, the line $AD$ lies between $AB$ and $AC$. So the $70^\circ$ angle at $A$ is split into $\angle BAD$ and the $20^\circ$ angle:
$$\angle BAD = 70^\circ - 20^\circ = 50^\circ.$$

The answer is $\boxed{\textbf{(D)}\ 50}$.

In an isosceles triangle, knowing the top angle tells you both bottom angles right away. When two isosceles triangles share the same base, compare their bottom angles at the same corner. A quick sketch shows whether you need the sum or the difference.

Both isosceles triangles have $AC$ as their base, so their base angles sit at $A$ and $C$.

In triangle $ABC$ with apex angle $40^\circ$, each base angle is
$$\angle BAC = \frac{180^\circ - 40^\circ}{2} = 70^\circ.$$
In triangle $ADC$ with apex angle $140^\circ$, each base angle is
$$\angle DAC = \frac{180^\circ - 140^\circ}{2} = 20^\circ.$$
Since $D$ lies inside triangle $ABC$, ray $AD$ lies between rays $AB$ and $AC$, so
$$\angle BAD = \angle BAC - \angle DAC = 70^\circ - 20^\circ = 50^\circ.$$

The answer is $\boxed{\textbf{(D)}\ 50}$.