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In the cube abcd-a1b1c1d1, M is the midpoint of dd1, O is the midpoint of AC. AB = 2. Verification: b1o ⊥ plane ACM. Can you tell me
Proof: connect b1o, b1m, OM
Because in the right angle △ bob1, Bo = √ 2, BB1 = 2, so b1o = √ 6
Because in the right angle △ md1b1, b1d1 = 2 √ 2, MD1 = 1, so b1m = √ 9 = 3
Because in the right angle △ MDO, OD = √ 2, MD = 1, so om = √ 3
Analysis of b1o, b1m and OM shows that they just form a right angle △ mb1o, and b1o ⊥ OM
Because △ ab1c is an isosceles triangle and O is the midpoint of AC, b1o ⊥ AC
Because om belongs to a line segment in plane ACM, b1o ⊥ plane ACM
In the cube abcd-a1b1c1d1, M is the midpoint of dd1 and O is the midpoint of AC
(1) connect Mo, connect B & # 185; O, connect B & # 185; m; (2) by combining two three lines, Mo is perpendicular to AC, B & # 185; O is perpendicular to AC; (3) so it is ∠ B & # 185; om dihedral angle; (4) by reverse Pythagorean, it is triangle B & # 185; in OM, ∠ B & # 185; OM is right angle; and (5) dihedral angle
In the cube abcd-a1b1c1d1, if M is the midpoint of dd1, we prove that BD1 is parallel to the plane Mac
Connect AC and BD, intersect at a point O, connect OM, M is the midpoint of dd1, O is the midpoint of BD, so OM is parallel to BD1,
Because the OM is in the plane Mac, BD1 is parallel to the plane Mac
In the quadrilateral ABCD, AC and BD intersect point O, and AC = BD, e and F are the midpoint of AD and BC respectively, and EF intersects AC and BD at point Mn respectively
It is proved that: let Q and R be the midpoint of AB and CD respectively, connecting EQ, QF, FR and re, the intersection of FR and AC is s, and the intersection of RE and BD is t. because e, Q, F and R are the midpoint of AD, AB, BC and CD respectively, QE ‖ BD is equal to 1 / 2bd, fr ‖ BD is equal to 1 / 2bd, QF ‖ AC is equal to 1 / 2Ac, er ‖ AC is equal to 1 / 2Ac, and AC =
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