(I) In\[\Delta \text{ }ABC\], we have
\[\angle ABC\text{ }=\text{ }2\angle ACB\][Given]
Presently, let \[\angle ACB\text{ }=\text{ }x\]
In this way, \[\angle ABC\text{ }=\text{ }2x\]
Likewise given, \[BP\] is bisector of \[\angle ABC\]
Subsequently, \[\angle ABP\text{ }=\angle PBC\text{ }=\text{ }x\]
By utilizing the point bisector hypothesis,
for example the bisector of a point isolates the side inverse to it in the proportion of other different sides.
Hence, \[CB:\text{ }BA\text{ }=\text{ }CP:\text{ }PA.\]
(ii) In \[\Delta \text{ }ABC\text{ }and\text{ }\Delta \text{ }APB,\]
\[\angle ABC\text{ }=\angle APB\][Exterior point property]
\[\angle BCP\text{ }=\angle ABP\] [Given]
Subsequently, \[\vartriangle ABC\text{ }\sim\text{ }\vartriangle APB\]by \[AA\]standard for comparability
Presently, since relating sides of comparative triangles are corresponding we have
\[CA/AB\text{ }=\text{ }BC/BP\]
Consequently, \[AB\text{ }x\text{ }BC\text{ }=\text{ }BP\text{ }x\text{ }CA\]