geometrytriangles
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?
If 1 and 2 are the same, then the angle bisect or theorem tells us that the third sides are equal, hence the triangle is isosceles.
Let $x$ be the common $A$-bisector/$B$-median/$C$-altitude.
In its role as the $A$-bisector, $x$ satisfies (by Stewart's Theorem and the Angle Bisector Theorem) $$b^2\,\frac{ac}{b+c}+c^2\,\frac{ab}{b+c}=a\left(x^2+\frac{a^2bc}{(b+c)^2}\right)\;\to\;x^2(b+c)^2=bc(a+b+c)(-a+b+c) \tag{1}$$
As the $B$-median (again by Stewart, or Apollonius), $$c^2\;\frac{b}{2}+a^2\;\frac{b}{2} = b \left(x^2+\frac{b^2}{4}\right)\quad\to\quad 4x^2 = 2a^2-b^2+2c^2 \tag{2}$$
As the $C$-altitude (invoking Heron's formula), $$\frac12cx=|\triangle ABC|\quad\to\quad4c^2x^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) \tag{3}$$
The non-linear system $(1)$, $(2)$, $(3)$ turns out to be quartic. Throwing it into Mathematica for numerical solution, and discarding non-positive values, gives two options:
$$(a,b,c,x) = \left(1,1,1,\frac12\sqrt{3}\right)\qquad (a,b,c,x) = \left(1, 1.2225\ldots, 0.2381\ldots, 0.3933\ldots\right) \tag{$\star$}$$
(The values in my comment to the question correspond to taking $c=1$.) The first solution is, of course, the equilateral triangle; the second is obtuse:
So, if we restrict ourselves to acute triangles, the equilateral becomes the only solution. $\square$
Best Answer
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: $$CF=8\sin A=8\sqrt{1-(2(\frac78)^2-1)^2}=8\sqrt{1-(\frac{17}{32})^2}=8\sqrt{1-\frac{17}{32}}\sqrt{1+\frac{17}{32}}=\frac74\sqrt{15}$$