For reference: Calculate the measure of the exterior bisector of angle "A" of a triangle ABC. If:$b-c = 20$ and $b\cdot c(p - b)(p - c)= 10c$ {$p$ ➔ semiperimeter) (Answer:$1$) *(It is possible that there is an error in the statement and it is $(b-c)(p-b)(p-c)$. It is not clear whether the sign is multiplication or subtraction in the original image of the question.)
My progress:
$b - c = 20\\ (bc)(p-b)(p-c)=10c$
Properties:
$AP^2 = mn - bc\\ \frac{AP}{m} = \frac{c}{n}\\ m=\frac{ab}{b-c}\\ n=\frac{ac}{c-b}\\ a^2 = mn-ac\\ AP = \frac{\sqrt{bc(p-b)(p-c)}}{|b-c|}\\ AP =\frac{ \sqrt{10c}}{20}\\ AP^2 = \frac{10c}{400}\\ AP^2 = \frac{c}{40}$
...?
By external angle bisector formula,
$ \displaystyle AP^2 = bc \left[ \left(\frac{a}{b-c}\right)^2 - 1\right]$
$ \displaystyle = \frac{bc}{400} \left[ a^2 - (b-c)^2 \right] = \frac{bc}{100} (p-b) (p-c)$
If $ ~bc (p-b)(p-c) = 100, AP = 1$. So I think it should read $100$ instead of $10c$. That seems to be the typo.