2007 AMC 12A Problems/Problem 6

The following problem is from both the 2007 AMC 12A #6 and 2007 AMC 10A #8, so both problems redirect to this page.

Problem

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 $BAD$ ?

$\mathrm{(A)}\ 20\qquad \mathrm{(B)}\ 30\qquad \mathrm{(C)}\ 40\qquad \mathrm{(D)}\ 50\qquad \mathrm{(E)}\ 60$

Solution 1

We angle chase and find out that:

$\angle DAC=\frac{180-140}{2} = 20$
$\angle BAC=\frac{180-40}{2} = 70$
$\angle BAD=\angle BAC- \angle DAC=50\ \mathrm{(D)}$

~minor edits by mobius247

Solution 2

Since triangle $ABC$ is isosceles we know that angle $\angle BAC = \angle BCA$ .

Also since triangle $ADC$ is isosceles we know that $\angle DAC = \angle DCA$ .

This implies that $\angle BAD = \angle BCD$ .

Then the sum of the interior angles of quadrilateral $ABCD$ is $40 + 220 + 2\angle BAD = 360$ .

Solving the equation, we get $\angle BAD = 50$ .

Therefore the answer is $\mathrm{(D)}$ .

Solution 3

Using the previous drawing we know that ACD and DAC are both equal to 20 (40/2 or (180-40)/2). We also know that BAC and BCA are both 70 or (180-40)/2. Thus BDA is 70-20 or 50. Easy. -RealityWrites

2007 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2007 AMC 12A Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

Solution 3

See also