In triangle ABC, one of the trisector of angle A is a median, and the other is an altitude. If BC = 24, what is the area of triangle ABC?
1. I've drawn a sketch (see attachment)
2. $\displaystyle \Delta(AMC)$ is an isosceles triangle and therefore the height h bisects half of $\displaystyle \overline{BC}$.
3. In the right triangle $\displaystyle \Delta(AMF)$ you know about the angle $\displaystyle \alpha$:
$\displaystyle \tan(\alpha)=\dfrac6h$
4. According to the question $\displaystyle \angle(BAM)=\alpha$.
In the right triangle $\displaystyle \Delta(ABF)$ you know:
$\displaystyle \tan(2\alpha)=\dfrac{18}h$
5. Since $\displaystyle \tan(2\alpha)=\dfrac{2\tan(\alpha)}{1-(\tan(\alpha))^2}$ you'll get:
$\displaystyle \dfrac{18}h = \dfrac{2\cdot \frac6h}{1-\left(\frac6h\right)^2}$
Solve for h and subsequently calculate the area of the triangle $\displaystyle \Delta(ABC)$
Spoiler:
I'm not certain that I understand your question correctly ...
1. This is a formula: $\displaystyle \tan(2\alpha)=\dfrac{2\tan(\alpha)}{1-\left(\tan(\alpha) \right)^2}$
2. In step #3 you find $\displaystyle \tan(\alpha)=\dfrac6h$
3. I then replaced $\displaystyle \tan(\alpha)$ in the formula by $\displaystyle \dfrac6h$
... or do you want me to derive the formula?