Creating an equation from this triangle.

Question

Pretty much stuck at where I should go. Any guidance?

Answer 1

Knowing the angle bisector theorem and denoting $BC = x$ you just need to note that $$\frac{6}{x}=\frac{4}{15-x}.$$ Solve for $x=BC$.

Answer 2

Use the law of sines in triangle ABD and also in triangle BDC; use what you know about $AB+BC$, what you know about angles ABD and DBC, and what you know about angles ADB and BDC.

Answer 3

Hints:

• Use the law of sines for each of triangles $\triangle ABD$ and $\triangle BDC$.

• You know that $AB + BC + CA = 25$ ( = perimeter), and

• $AC = 4 + 6 = 10$, so $AB + BC = 25 - 10 = 15$

• You know that $\angle ABD \cong \angle CBD$, and the $\angle ADB + \angle BDC = 180^\circ$ (as they are are supplementary).

You have all the information you need to solve for $BC$