As shown in the figure, in the pyramid s-abcd, the bottom surface ABCD is a parallelogram, the side SBC ⊥ the bottom surface ABCD, ∠ ABC = 45 °, SA = sb, proving that SA ⊥ BC

As shown in the figure, in the pyramid s-abcd, the bottom surface ABCD is a parallelogram, the side SBC ⊥ the bottom surface ABCD, ∠ ABC = 45 °, SA = sb, proving that SA ⊥ BC

It is proved that: for so ⊥ BC, the perpendicular foot is O, connecting Ao, so, ⊙ bottom ABCD is parallelogram, side SBC ⊥ bottom ABCD, side SBC ∩ bottom ABCD = BC, ⊥ so ⊥ bottom ABCD, and ⊂ OA ⊂ bottom ABCD, ob ⊂ bottom ABCD, ⊥ so ⊥ OA, so ⊥ ob, and & nbsp; SA = sb, ∩ OA = ob, ABC = 45 °, OA ⊥ ob, ≁ BC ⊥ so, BC ⊥ Ao, so ∩ Ao = O, ∩ BC ⊥ planar SOA, SA ⊂ planar SOA ≁ SA ⊥ BC