The diagram is given with coordinates
[tex]A(-a,0),P(0,b),B(a,0)[/tex]To find the value of AP,
[tex]AP=\sqrt[]{(0-(-a))^2+(b-0)^2}[/tex][tex]AP=\sqrt[]{a^2+b^2}[/tex]To find the value of BP.
[tex]BP=\sqrt[]{(a-0)^2+(0-b)^2}[/tex][tex]BP=\sqrt[]{a^2+b^2}[/tex]Hence it is proved that AP=BP.