In triangle ABC, the angle bisector of angle BAC meets AC at D. If angle BAC = 60, angle ABC = 60, and AD = 24, then find the area of triangle ABC.
Since angle BAC is 60 degrees and angle ABC is 60 degrees, we know that angle ACB is also 60 degrees, making triangle ABC an equilateral triangle.
Let E be the foot of the perpendicular from point B to side AC, and let x be the length of BE. Then, the length of EC is also x, and the length of AB is 2x.
Since AD is the angle bisector of angle BAC, we know that the ratio of the length of AC to the length of AB is equal to the ratio of the length of DC to the length of BD:
AC / AB = DC / BD
Substituting the values we know, we get:
1 / 2 = DC / (x + 24)
Solving for DC, we get:
DC = (x + 24) / 2
Using the Pythagorean theorem, we can find x in terms of DC:
x^2 = AB^2 - AE^2 = (2x)^2 - DC^2 = 4x^2 - (x + 24)^2 / 4
Simplifying this equation, we get:
15x^2 - 12x - 576 = 0
Solving for x using the quadratic formula, we get:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
where a = 15, b = -12, and c = -576. Taking the positive root, we get:
x = 12
Therefore, the length of AB is 2x = 24, and the area of triangle ABC is:
Area = (sqrt(3) / 4) AB^2 = (sqrt(3) / 4) (24)^2 = 144sqrt(3)
Therefore, the area of triangle ABC is 144sqrt(3).