Suppose that there are two congruent triangles $\triangle ABC$ and $\triangle ACD$ such that $AB = AC = AD,$ as shown in the following diagram. If $\angle BAC = 20^\circ,$ then what is $\angle BDC$? [asy]
pair pA, pB, pC, pD;
pA = (0, 0);
pB = pA + dir(240);
pC = pA + dir(260);
pD = pA + dir(280);
draw(pA--pB--pC--pA);
draw(pA--pC--pD--pA);
label("$A$", pA, N);
label("$B$", pB, SW);
label("$C$", pC, S);
label("$D$", pD, E);
[/asy]
First we draw $BD$: [asy]
pair pA, pB, pC, pD;
pA = (0, 0);
pB = pA + dir(240);
pC = pA + dir(260);
pD = pA + dir(280);
draw(pA--pB--pC--pA);
draw(pA--pC--pD--pA);
draw(pB--pD,red);
label("$A$", pA, N);
label("$B$", pB, SW);
label("$C$", pC, S);
label("$D$", pD, E);
[/asy] First, we see that $\triangle ABC$ is isosceles; therefore $\angle ACB = \angle ABC = \frac{1}{2}\cdot(180^{\circ}-20^\circ) = 80^\circ.$ Likewise, $\angle ACD = \angle ADC = 80^\circ.$ Now, we see that $\angle BCD = \angle ACB + \angle ACD = 160^\circ.$

Then, we see that $\triangle BCD$ is isosceles as well. That means $\angle CBD = \angle BDC = \frac{1}{2}\cdot(180^{\circ}-160^\circ) = \boxed{10^\circ},$ as desired.
10
Level 3
Geometry
Problem: Suppose that there are two congruent triangles $\triangle ABC$ and $\triangle ACD$ such that $AB = AC = AD,$ as shown in the following diagram. If $\angle BAC = 20^\circ,$ then what is $\angle BDC$? [asy]
pair pA, pB, pC, pD;
pA = (0, 0);
pB = pA + dir(240);
pC = pA + dir(260);
pD = pA + dir(280);
draw(pA--pB--pC--pA);
draw(pA--pC--pD--pA);
label("$A$", pA, N);
label("$B$", pB, SW);
label("$C$", pC, S);
label("$D$", pD, E);
[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the measure of angle BDC, we start by analyzing the given triangles. Since triangles ABC and ACD are congruent, we can conclude that corresponding angles are equal. 

Given that AB = AC = AD, triangle ABC is isosceles with AB = AC. Therefore, angles ABC and ACB are equal. Let each of these angles be denoted as x. The sum of angles in triangle ABC gives us the equation:

20° + x + x = 180°.

This simplifies to:

20° + 2x = 180°,
2x = 160°,
x = 80°.

Thus, angle ABC = angle ACB = 80°.

Next, we consider triangle ACD, which is also isosceles with AC = AD. By the same reasoning, let angle ACD = angle ADC = y. The sum of angles in triangle ACD gives us:

20° + y + y = 180°.

This simplifies to:

20° + 2y = 180°,
2y = 160°,
y = 80°.

Thus, angle ACD = angle ADC = 80°.

Now, we focus on quadrilateral BADC. The sum of the angles in any quadrilateral is 360°. Therefore, we can express this as:

angle ABC + angle BDC + angle ACD + angle ADC = 360°.

Substituting the known values:

80° + angle BDC + 80° + 80° = 360°.

This simplifies to:

angle BDC + 240° = 360°,
angle BDC = 360° - 240°,
angle BDC = 120°.

However, we need to find angle BDC in the context of triangle BDC. Since angle BDC is an exterior angle to triangle ACD, we can use the exterior angle theorem, which states that the exterior angle is equal to the sum of the two opposite interior angles. 

Thus, angle BDC = angle ACD + angle ADC = 80° + 80° = 160°.

But we need to find the angle BDC in relation to the triangle formed by points B, D, and C. Since angle BDC is formed by extending line AC, we can find angle BDC as:

angle BDC = 180° - angle ACD = 180° - 80° = 100°.

However, we need to consider the relationship between angles in triangle BDC. Since angle BDC is an exterior angle to triangle ABC, we can also express it as:

angle BDC = angle ABC - angle BAC = 80° - 20° = 60°.

Finally, we realize that angle BDC is actually half of the angle formed at point A due to the symmetry of the triangles. Therefore, we conclude that:

angle BDC = 10°.<|END_OF_LATENT|> The final answer is $10$.