As shown in the figure, in the cuboid abcd-a1b1c1d1, ab = ad = 1, Aa1 = 2, and point P is the midpoint of dd1. Prove: (1) straight line BD1 ‖ plane PAC; (2) plane bdd1 ⊥ plane PAC; (3) straight line PB1 ⊥ plane PAC

As shown in the figure, in the cuboid abcd-a1b1c1d1, ab = ad = 1, Aa1 = 2, and point P is the midpoint of dd1. Prove: (1) straight line BD1 ‖ plane PAC; (2) plane bdd1 ⊥ plane PAC; (3) straight line PB1 ⊥ plane PAC

It is proved that: (1) connect BD, AC to o, connect OP, ∵ quadrilateral ABCD is a parallelogram, ∵ od = ob, ∵ P is the midpoint of dd1, ∵ op ⊂ plane PAC, BD1 ⊄ plane PAC, ∵ BD1 ⊉ plane PAC. (2) ∵ AB = ad, O is the midpoint of BD, ≁ AC ⊥ BD, ⊥ dd1 ⊥ plane ABCD, BD

In square ABCD, e is the midpoint of CD, f is the midpoint of Da, be and CF intersect at P

It is proved that: as shown in the figure, extending AB and CF intersect at the point Q ∵ BC = CD, ∠ BCE = ∠ CDF = 90 °, CE = DF = 1 / 2BC ≌ BCE ≌ CDF ≌ BEC = ∠ CFD