2009 AMC 10B Problem 9

The equal sides make two isosceles triangles, one on each side of $C$. Solve the left one first, because we are told how its angles compare.

1. In triangle $ABC$, sides $AB$ and $BC$ are equal. The angles opposite equal sides are equal, so $\angle A = \angle BCA$. (The angle at $B$ sits between the two equal sides, so it is the odd one out.)
2. Call $\angle B = x$. Then $\angle A = \angle BCA = \frac52 x$.
3. The angles of a triangle add to $180^\circ$:
$$\tfrac52 x + \tfrac52 x + x = 6x = 180^\circ,$$
so $x = 30^\circ$ and $\angle BCA = \frac52 \times 30^\circ = 75^\circ$.
4. Lines $AE$ and $BD$ cross at $C$. The angle $\angle DCE$ is directly across from $\angle BCA$ (vertical angles), so it is also $75^\circ$.
5. In triangle $CDE$, sides $CD$ and $CE$ are equal, so $\angle D = \angle E$. These two angles share what is left of $180^\circ$:
$$\angle D = \frac{180^\circ - 75^\circ}{2} = \frac{105^\circ}{2} = 52.5^\circ.$$

The answer is $\boxed{\textbf{(A)}\ 52.5}$.

Equal segments mean isosceles triangles, and in an isosceles triangle one angle tells you the other two. The crossing point $C$ passes an angle from one triangle to the other, because vertical angles are equal. The one thing to get right is which angle sits between the two equal sides; the other two angles are the matching pair.

In triangle $ABC$, the sides $AB$ and $BC$ are equal, so the base angles at $A$ and $C$ are equal. Let $\angle B = x$; then $\angle A = \angle BCA = \frac52 x$, and the angle sum gives
$$\frac52 x + \frac52 x + x = 6x = 180^\circ, \qquad x = 30^\circ.$$
So $\angle BCA = 75^\circ$.

Since $A, C, E$ are collinear and $B, C, D$ are collinear, $\angle DCE$ is vertical to $\angle BCA$, hence also $75^\circ$.

In triangle $CDE$, the sides $CD$ and $CE$ are equal, so the base angles at $D$ and $E$ are equal:
$$\angle D = \frac{180^\circ - 75^\circ}{2} = 52.5^\circ.$$

The answer is $\boxed{\textbf{(A)}\ 52.5}$.