Answer:
\[A{{C}^{2}}=B{{C}^{2}}-A{{B}^{2}}\] [Given] \[A{{C}^{2}}+A{{B}^{2}}=B{{C}^{2}}\] \[\therefore \angle BAC=90{}^\circ \] [By converse of Pythagoras? theorem] \[\Delta \,APB\sim \Delta \,CPA\] \[\Rightarrow \frac{AP}{CP}=\frac{PB}{PA}\] [In similar triangle, corresponding sides are proportional] \[\Rightarrow P{{A}^{2}}=PB.CP\] Hence Proved.
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