Knowing the length of two sides of a triangle and the angle bisector in between , find the length of one of the altitude.

Question

In $\triangle ABC$, $AB = 6, AC = 8$ and internal angle bisector $AD = 6$ such that $D$ lies on segment $ BC$. Compute the length of altitude $CF$ where $F$ is a point on line $AB$.

For calculating $CF$ , we will need area of the triangle.
For calculating area , we will need $BC$ .( Then we can use heron's formula)
How can I calculate $BC$?
Also it is given that , the angle bisector , $AD$ is $6$.
How can I utilize this information?

Answer 1

Summing up the area of two triangles we get, $3\sin (\frac{A}{2})+4\sin (\frac{A}{2})=7\sin (\frac{A}{2})=4\sin A$. Thus, $\cos (\frac{A}{2})=\frac78$. Or $\cos A=2(\frac78)^2-1$.

So: $$CF=8\sin A=8\sqrt{1-(2(\frac78)^2-1)^2}=8\sqrt{1-(\frac{17}{32})^2}=8\sqrt{1-\frac{17}{32}}\sqrt{1+\frac{17}{32}}=\frac74\sqrt{15}$$

Answer 2

Sorry to say that your diagram is misleading. If AB = AD = 6, then it should look like:-

AH is another altitude of triangle ABC.

If angle BAH = x, then angle CAH = 3x

In triangle ABH, setup a relation between (6, AH, x)

In triangle ACH, setup a relation between (8, AH, 3x)

Eliminating AH from the two, you will get cos(3x) : cos x = 6 : 8

Use compound (and also double) angle formulas to get x.

Once x is known, the rest is easy.

Answer 3

You can find the lengh of the bisector using this formula(if you need I can put the proof)

here is the solution: 36=48*(1-x^2/194) then we have x=6.964194.