question_answer

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

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.