Pyramid SABC has right triangular base ABC, with ∠ A B C = 90 ∘...

Step 1
Let K be a middle point of AC.
Just $\measuredangle SBK=\measuredangle SAK=\arccos \frac{\sqrt{3}}{2}={30}^{\circ }.$.
BK is a median of $\mathrm{\Delta }ABC$ and is not perpendicular to AC, otherwise $AB=BC$, which is a contradiction.
By the way, $BK\perp SK$, but we said about it in the first line.
Step 2
Let SK′ be an altitude of the pyramid.
Thus, since $SA=SB=SC$, we obtain: ΔSAK′≅ΔSBK′ and $\mathrm{\Delta }SA{K}^{\prime }\cong \mathrm{\Delta }SC{K}^{\prime }$, which gives $A{K}^{\prime }=B{K}^{\prime }=C{K}^{\prime },$, which says K′ is a center of the circumcircle for $\mathrm{\Delta }ABC$.
Thus, $K^{\prime }\equiv K$.