Finding $\overline {AD} \times \overline {CD}$ in $\triangle APB$

[Albert1]

$\triangle APB, \overline {PA}=\overline {PB}, \angle APB=2\angle ACB, $
point $D$ is the intersection of $\overline {AC}$, and $\overline {BP}$
if $\overline {BP}=3 , \overline {PD}=2$
please find the value of $\overline {AD}\times \overline {CD}$

 

[Greg]

Hi Albert. Where is point $C$?

 

[RLBrown]

Spoiler

Since the problem implies (truthfully) that it does not matter where C or A is located up to the constraints (on circle), I picked AC to be vertical.

View attachment 4989

This depicts one of the Pythagorean Means, specifically the Geometric Mean.
That is to say; Length AD is the GM of DB and 6-DB. AD = SQRT(1*5).

AD=DC so AD*DC = 5
(proof of the implication not being required, makes this a simple problem)

 

[Albert1]

Spoiler

Since the problem implies (truthfully) that it does not matter where C or A is located up to the constraints (on circle), I picked AC to be vertical.
This depicts one of the Pythagorean Means, specifically the Geometric Mean.
That is to say; Length AD is the GM of DB and 6-DB. AD = SQRT(1*5).

AD=DC so AD*DC = 5
(proof of the implication not being required, makes this a simple problem)

very good solution !