• # question_answer In triangle $ABC$, if $AP\bot BC$ and $A{{C}^{2}}=B{{C}^{2}}-A{{B}^{2}}$, then prove that $P{{A}^{2}}=PB\times CP$.

 $A{{C}^{2}}=B{{C}^{2}}-A{{B}^{2}}$    [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.