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2009 AMC 10B Problem 9

2009 AMC 10B Problem 9AMC 10 Step by Step

Name: Date: September 13, 2026

Angles & Polygons · difficulty 2 of 5 · about 3 min · position P1-10

Segment $BD$ and $AE$ intersect at $C$ , as shown, $AB=BC=CD=CE$ , and $\angle A = \frac 52 \angle B$ . What is the degree measure of $\angle D$ ?

A)
$52.5$
B)
$55$
C)
$57.5$
D)
$60$
E)
$62.5$

Answer

Key insight

Triangle ABC is isosceles with apex B, so its angles are 5x/2, 5x/2, x giving x = 30; vertical angles then pass 75 degrees to isosceles triangle CDE.

Solution

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 $\boxed{\textbf{(A)}\ 52.5}$ .

Why this works

Chains of equal segments create isosceles triangles, and each isosceles triangle converts one known angle into the other two. The intersection point $C$ passes information across via vertical angles. Identify which vertex is the apex (between the two equal sides) before assigning base angles.

The trap

Assuming angle B is the base angle of triangle ABC rather than the apex angle between the equal sides AB and BC.

Common mistakes

Assuming angle B is the base angle of triangle ABC rather than the apex angle between the equal sides AB and BC.
Using the supplementary angle $105^\circ$ at $C$ for triangle $CDE$ instead of the vertical angle $75^\circ$ .

Techniques

Set up the equation/formula and compute; no special trick needed