In a triangle $\triangle ABC$, the bisector of angle from the point $A$ intersects $\overline {BC}$ in point $D$. Prove: $|AD|^2=|AB|\cdot |AC|-|DB|\cdot |DC|$.
I don't even know where to start. I'll probably have to use the power of a point on a circle theorem but I don't know how.
You can use the Steward's Theorem to get:
$$AB^2\cdot DC + AC^2 \cdot DB = BC(DC \cdot DB + AD^2)$$
$$AD^2 = \frac{AB^2\cdot DC + AC^2 \cdot DB}{BC} - DC \cdot DB$$
So we need to prove that: $\frac{AB^2\cdot DC + AC^2 \cdot DB}{BC} = AB \cdot AC$. For this we can use the Angle Bisector Theorem. So:
$$AB^2 \cdot DC + AC^2 \cdot DB = AB \cdot \frac{AC}{DC} \cdot DB \cdot DC + AC \cdot \frac{AB}{BD} \cdot DC \cdot BD$$ $$ = AC \cdot AB (DC + BD) = AC \cdot AB \cdot BC$$
Thus:
$$\frac{AB^2\cdot DC + AC^2 \cdot DB}{BC} = \frac{AC \cdot AB \cdot BC}{BC} = AC \cdot AB$$
Hence the proof.