2007 AMC 10A Problem 8

P1-10~2 minPrint

Angles & Polygons Triangles: Area & Pythagorean Set up the equation/formula and compute; no special trick needed

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 $BAD$ ?

Show solution without answering

Key insight

Both isosceles triangles have AC as the base. Find the base angle at A in each one (70 and 20 degrees); angle BAD is the gap between them.

Solution

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 2: Angles
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 9: Polygons (§9.1, §9.2)
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)