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?

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- Feb 18th 2011, 12:59 AMKalodaangle trisector?
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?

- Feb 18th 2011, 05:06 AMearboth
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__: - Feb 23rd 2011, 05:13 PMfireballs619
I was looking over this, and am somewhat confused about step 5 the solution. I am not sure how you got the second half of that fraction. If possible, could you elaborate on that? Thanks!

- Feb 23rd 2011, 10:51 PMearboth
I'm not certain that I understand your question correctly ... (Thinking)

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? - Feb 24th 2011, 04:16 AMfireballs619
Oh, my apologies, as I did not realize that it was a formula. I thought you had somehow derived it from the diagram, and I could not for the life of me figure out how. Thanks for the time, though!