$ABC$ is a triangle inscribed in a circle. $AD, AE$ are straight lines drawn from the vertex A to the base BC parallel to the tangents at B and C respectively. If $AB = 5$ cm, $AC = 6$ cm, and $CE = 9$ cm, then what is the length of $BD$ (in cm)?
I tried using cosine rule. But I am stuck anyways.
Rename $E$ and $D$ in text of OP. We see that $a+x=9$.
Since we have $$\angle BDA = \angle BAC$$ wesee that $AC$ is tangent on circle $ABD$ so we have $$CA^2 = CB\cdot CD \implies 36 = 9\cdot a \implies a=4$$
Similary we have $$BA^2 = BE\cdot BC \implies 25 = (a+y)a \implies y=2.25$$