+359 In triangle $ABC,$ the angle bisector of $\angle BAC$ meets $\overline{BC}$ at $D.$ If $\angle BAC = 60^\circ,$ $\angle ABC = 60^\circ,$ and $AD = 24,$ then find the area of triangle $ABC.$
+399 
Because BAC = 60, ABC = 60, ACB should equal 60.
Therefore, ABC is equilateral.
Because AD is an angle bisector, then ABD, is 30, and so ADB = ADC = 90.
Therefore, set BD = x.
\({x}^{2}+{24}^{2}={(2x)}^{2}\).
We get \(x = 8\sqrt{3}\). (These triangle ratios are very special, see if you can find the ratios for a 30-60-90 triangle for yourself)
Therefore the area is \(\frac{8\sqrt{3}*24*2}{2}=192\sqrt{3}\).
+399
Because BAC = 60, ABC = 60, ACB should equal 60.
Therefore, ABC is equilateral.
Because AD is an angle bisector, then ABD, is 30, and so ADB = ADC = 90.
Therefore, set BD = x.
\({x}^{2}+{24}^{2}={(2x)}^{2}\).
We get \(x = 8\sqrt{3}\). (These triangle ratios are very special, see if you can find the ratios for a 30-60-90 triangle for yourself)
Therefore the area is \(\frac{8\sqrt{3}*24*2}{2}=192\sqrt{3}\).
