(1) To prove AF⊥SC, we only need to know the vector AF⊥ SC. Vector AF= vector AE+ vector EF. Vector autofocus? Vector SC=|AF|? |SC|？ Cos< angle between two vectors > = (vector AE+ vector EF)? Vector SC

Vector AE? Vector SC=0, because AE is perpendicular to SBC plane, vector EF? Vector SC=0, because EF⊥SC. The result of the above formula is zero, that is to say, the cosine of the angle between vector AF and vector SC is 0, that is, 90 degrees.

(2) Still use vectors. In order to prove AG⊥SD, we only need to know the vector AG⊥ vector SD. Vector SD= vector SC+ vector CD. Vector AG？ Vector SD=|AG|? |SD|？ Cos< Angle between two vectors > = (vector SC+ vector CD)? Vector AG= vector SC? Vector AG+ vector CD? Vector stock company. Since EF⊥SC and AF⊥SC are perpendicular to AEF, SC⊥AG, that is, the first term in the above formula, is zero. And vector CD Vector AG= vector CD? Vector AS+ vector CD? Vector SG, CD⊥AS, CD⊥SG, the second term of the above formula is also zero. Vector AG？ The vector SD=0 and the included angle is 90 degrees.