The points $(-8,0),(8,0),(0,5)$ are the vertices of a  

$AC=\sqrt{{(-8-0)}^2+{(0-5)}^2}=\sqrt{64+25}=\sqrt{89}$

$BC=\sqrt{{(8-0)}^2+{(0-5)}^2}=\sqrt{64+25}=\sqrt{89}$

$AB=\sqrt{{(-8-8)}^2+{(0-0)}^2}=\sqrt{256}=16$

$AC=BC\ne AB$

$\therefore \Delta ABC$ is an isosceles triangle.