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2007 AMC 10A Problem 8

2007 AMC 10A Problem 8AMC 10 Step by Step

Name: Date: September 13, 2026

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

Triangles $ABC$ and $ADC$ are isosceles with $AB=BC$ and $AD=DC$ . Point $D$ is inside triangle $ABC$ , angle $ABC$ measures $40$ degrees, and angle $ADC$ measures $140$ degrees. What is the degree measure of angle $BAD$ ?

A)
$20$
B)
$30$
C)
$40$
D)
$50$
E)
$60$

Answer

Key insight

Both triangles share base AC, so angle BAD is the difference of the two base angles: 70 - 20 = 50.

Solution

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 $\boxed{\textbf{(D)}\ 50}$ .

Why this works

An isosceles triangle's apex angle determines its base angles instantly. When two isosceles triangles share a base, their apexes lie on the same perpendicular bisector, and the angle between the two equal sides at a base vertex is simply the difference of the base angles. Sketching the figure fixes which angle is a sum and which is a difference.

The trap

Adding the base angles (70 + 20 = 90) or misplacing D outside the triangle, which changes which angle is a difference.

Common mistakes

Adding the base angles ( $70 + 20 = 90$ ) or misplacing $D$ outside the triangle, which changes which angle is a difference.
Treating $40^\circ$ or $140^\circ$ as a base angle rather than the apex angle.

Techniques

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