"the projection of point a on the plane sbc is the outer center of the triangle sbc",
The projection of point a in the plane bsc is d,
So: point D is the outer center of triangle sbc (that is, point D is the intersection of the perpendicular lines of BS, SC and bc).
∵? "D is at the midpoint of the hypotenuse BC of the triangle bsc" and AB=AC,
∴△ABC is an isosceles triangle, so AD is perpendicular to BC and sd = cd = bd.
AD=AD,
AS=AB,
DS=DB,
(SSS),
So △ ADB △ ads,
So ∠ADB=∠ADS=? 90 ,
AD⊥BC,AD⊥DS,
BC and DS are on BSC plane,
BC and DS intersect at point d,
∴AD⊥ Bachelor of Science in Plane.
I don't know if it's right. Ha ha. . ? I drew another picture myself.