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 $AB$. figure

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?

  • $\begingroup$ I am getting an answer of (7/4)*sqrt(15). Since its looks ugly, I fear it may not be correct, is it so? $\endgroup$ – Sawarnik Apr 4 '14 at 18:05
  • $\begingroup$ @Sawarnik Answer is correct ! you're awesome . So please post the solution $\endgroup$ – A Googler Apr 4 '14 at 18:06

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

  • 1
    $\begingroup$ I don't think it is ugly. It is quite nice! Edit: it is correct now $\endgroup$ – A Googler Apr 4 '14 at 18:53

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

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.


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

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

  • $\begingroup$ Hi Taha, welcome to MathsSE! As it stands the answer is not complete. You do not need the proof, but at least you could explain how to use this formula and define the symbols. $\endgroup$ – Andrea May 25 '16 at 11:25

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