Finding vector form of an angle bisector in a triangle

Question

Find vector form of angle bisector, $\vec{BP}$, using $\vec{b}$ and $\vec{c}$.

That's how far I've got. Please don't use $tb+ (1-t)b$, or similar since I don't know what that is. Just basic dot product, vector product, triple product...If possible.

Answer 1

$$\overrightarrow{AP}=k\left(\frac{\underline{c}}{|c|}+\frac{\underline{b}}{|b|}\right)$$ $$\Rightarrow\overrightarrow{BP}=-\underline{c}+k\left(\frac{\underline{c}}{|c|}+\frac{\underline{b}}{|b|}\right), \text{and}$$ $$\Rightarrow\overrightarrow{PC}=\underline{b}-k\left(\frac{\underline{c}}{|c|}+\frac{\underline{b}}{|b|}\right)$$ $$\Rightarrow\overrightarrow{BP}\text{ is parallel to }\overrightarrow{PC}\Rightarrow-\underline{c}+k\left(\frac{\underline{c}}{|c|}+\frac{\underline{b}}{|b|}\right)=\alpha\left(\underline{b}-k\left(\frac{\underline{c}}{|c|}+\frac{\underline{b}}{|b|}\right)\right)$$ Since $\underline{b}$ and $\underline{c}$ are independent, we can equate coefficients and solve simultaneously to find $k.$

After a couple of lines, we get $$k=\frac{|b||c|}{|b|+|c|}$$

Answer 2

Since $|\vec{c}|$ / $|\vec{BP}|$= $|\vec{b}|$ / $|\vec{CP}|$ we can express length of $|\vec{BP}|$ using familiar vectors, and then just add the vectors.