Question

Let ABC be a triangle.  Let D be a point on side \overline{AC} such that line segment \overline{BD} bisects \angle ABC.  If \angle A = 30^\circ, \angle C = 80^\circ, and AC = 12, then find the area of triangle ABD.

Give your answer in exact form, using radicals as necessary. You can write in \LaTeX, or for example write sqrt(2) to indicate \sqrt 2.

Answer 1

[AI-generated answer — Bot GPT 3.5]

Let E be the foot of the perpendicular from B to AC. Since B is the internal angle bisector, we have $\angle ABD = \angle DBC = \frac{1}{2} \times 30 = 15$. Therefore, by using angles in triangle ABD, we have $\angle ADB = 180 - 30 - 15 = 135$. Likewise, using angles in triangle BDC, we have $\angle DBC = \angle DCB = 80/2 = 40$. Therefore, $\angle BDE = 180 - 90 - 40 = 50$.

In triangle BDE, $\frac{BD}{\sin 50} = \frac{12}{\sin 40} \implies BD = \frac{12\sin 50}{\sin 40}$. Using the sine rule in triangle ABD, we get $AB = \frac{BD\sin 15}{\sin 135} = \frac{12\sin 50 \cdot \sin 15}{\sin 40}$.

Therefore, the area of triangle ABD is $\frac{1}{2} \times AB \times BD = \frac{1}{2} \times \frac{12\sin 50 \cdot \sin 15}{\sin 40} \times \frac{12\sin 50}{\sin 40} = 72\sin 50 \cdot \sin 15$. Therefore, the area of triangle ABD is $\boxed{18\sin 150 = 18\sin 30 = 18\times \frac{1}{2} = 9}$.