Three questions about solid geometry in senior high school, 1. In cube AC1, e is the point on edge CC1, and a = C1E / EC, (1) if plane bed1 ⊥ plane bdd1b1, then a=_____ (2) if bed1 ⊥ ab1c, then a=_____ . 2. (1) in the diamond ABCD, a = 60 ° AB = 4, after folding it into a straight dihedral angle along BD, AC = 4=_____ The tangent of dihedral angle a-cd-b is equal to_____ . (2) In rectangle ABCD, ab = 3, ad = 4, it is folded into a straight dihedral angle along BD, AC=_____ . 3. It is known that ABCD is a rectangle, PA ⊥ plane ABCD, M is the midpoint of PC, PA = ad (1) Verification: plane mAb ⊥ plane PCD (2) Finding the size of dihedral angle m-ab-c

Three questions about solid geometry in senior high school, 1. In cube AC1, e is the point on edge CC1, and a = C1E / EC, (1) if plane bed1 ⊥ plane bdd1b1, then a=_____ (2) if bed1 ⊥ ab1c, then a=_____ . 2. (1) in the diamond ABCD, a = 60 ° AB = 4, after folding it into a straight dihedral angle along BD, AC = 4=_____ The tangent of dihedral angle a-cd-b is equal to_____ . (2) In rectangle ABCD, ab = 3, ad = 4, it is folded into a straight dihedral angle along BD, AC=_____ . 3. It is known that ABCD is a rectangle, PA ⊥ plane ABCD, M is the midpoint of PC, PA = ad (1) Verification: plane mAb ⊥ plane PCD (2) Finding the size of dihedral angle m-ab-c

1. (1) when a = 1, that is, e is the midpoint of CC1, plane bed1 ⊥ plane bdd1b1. Connecting BD1 and b1d, the intersection point is o. in cube BD1, point O bisects BD1 and b1de. When BD1 and b1de are the midpoint of CC1, ED1 = be. In isosceles triangle ed1b, EO is the middle line ⊥ EO ⊥ BD1 on the bottom BD1. Similarly, EO ⊥ b1d ⊥ EO ⊥ bb1d1d1d plane, then plane be

In solid geometry, three planes intersect in a straight line. If the size of two dihedral angles is known, can we directly use the addition and subtraction method to find another angle?

First prove that the line belongs to three planes at the same time, and then subtract the other two known planes with 180
If it can be proved that the intersecting lines of three planes are parallel to each other, the other two known planes can also be subtracted by 180

If PA = Pb = PC, then o is ()
A. Center of gravity

∵ P is the point outside the plane where △ ABC is located, and point O is the projection of point P on plane ABC, and ∵ PA = Pb = PC, then the distance from point O to a, B, C is equal, that is, OA = ob = OC, then point O is the outer center of △ ABC, so a is selected

If the lengths of the three sides of △ ABC are 3, 4 and 5 respectively, P is a point outside the plane ABC, the distance from it to the three sides is equal to 2, and the projection o of P on the plane ABC is located inside △ ABC, then Po is equal to ()
A. 1B. 2C. 32D. 3

In RT △ pod, RT △ Poe, RT △ POF, Po is common, OD = OE = of can be obtained by Pythagorean theorem, O should be the heart of △ ABC. Let od = R, in RT △ ABC, according to the area, we can get 12 × 3 × 4 = 12R × (3 + 4 + 5), the solution is r = 1. In RT △ pod, Po = Pd2 − OD2 = 22 − 12 = 3

It is known that PA and Pb are tangent lines of circle O, a and B are tangent points, ∠ P = 60 degrees, PA = 6 √ 3. Find the length of arc ab

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If AB = 8, the chord center distance of AB is 3, then the length of PA is ()
A. 5B. 203C. 253D. 8

Connecting OA, ob, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\therefore, it can be concluded that the formula is 5 × 43 = 203 Choose B

As shown in the figure, PA and Pb are tangent lines of circle O, a and B are tangent points, P is intersection point of extension line of tangent points a and B, angle P = 60 degrees, ab = 6 root sign 3, and radius of circle O is calculated

∠AOB=180°-∠P=120°
In the triangle AOB, according to the cosine theorem:
(6√3)^2=r^2+r^2-2r^2cos120°
3r^2=36*3
r^2=36
r=6

As shown in the figure: Mn is the tangent line of ⊙ o, a is the tangent point, through point a, make the chord of AP ⊥ Mn intersection ⊙ o, BC at point P, if PA = 2cm, Pb = 5cm, PC = 3cm. Find the diameter of ⊙ o

Extending AP intersection ⊙ o at point D; ∵ PA · PD = PC · Pb, ∵ 2 × PD = 3 × 5, ∵ PD = 7.5cm, ∵ o diameter ad = PA + PD = 2 + 7.5 = 9.5cm

As shown in the figure, in ⊙ o, P is any point of diameter Mn, through P, make chord AC and BD, so that ⊙ APN = ⊙ BPN, verify PA = Pb
Both APN and BPN are obtuse angles

Respectively through p
If O is OE ⊥ AC in E, of ⊥ BD in F, then ∠ OFP = ∠ OEP = 90 °, AE = 1 / 2Ac, BF = 1 / 2bd
∵∠APN=∠BPN,∠APD=∠BPC
∴∠FPO=∠EPO
∵PO=PO
∴△EPO≌△FPO
∴OE=OF,PE=PF
∴AC=BD
∴AE=BF
∴AE-PE=BF-PF
That is pa = Pb

In the circle O, AB is the diameter, P is the point on AB, passing through P is the chord Mn, angle NPB = 45 degrees (1) if AP = 2, BP = 6, find the length of Mn (2) if MP = 3
In the circle O, AB is the diameter, P is the point on AB, passing through P is the chord Mn, angle NPB = 45 degrees (1) if AP = 2, BP = 6, find the length of Mn (2) if MP = 3, NP = 5, find the length of ab (3) if the radius of circle O is r, find the square of PM + the square of PN

Because the diameter AB = AP + BP = 2 + 6 = 8, the radius OA = 8 / 2 = 4, Op = oa-ap = 4-2 = 2. And the angle MPB = 45 degrees, so oh is vertical Mn, the perpendicular foot is h, the triangle OHP is isosceles right triangle. Oh = HP, and oh ^ 2 + pH ^ 2 = OP ^ 2, so oh = pH = op / (radical 2) = radical 2. Then, the vertical chord passing through the center bisects the chord, so MH = NH, connecting OM. In the right triangle ohm, using the Pythagorean theorem, MH ^ 2 = Mo ^ 2-OH ^ 2 = 4 ^ 2 - (radical 2) ^ 2 = 14, MH = radical 14, Therefore, MP = radical 14 - radical 2, NP = radical 14 + radical 2, Mn = 2, radical 14.2 if MP = 3, NP = 5, then Mn = 3 + 5 = 8, MH = 8 / 2 = 4, pH = 1
3 because MH = NH, oh = HP, oh is vertical Mn, then PM ^ 2 + PN ^ 2 = (mh-hp) ^ 2 + (Ph + HN) ^ 2 = (mh-hp) ^ 2 + (MH + Hp) ^ 2 = 2 (MH ^ 2 + Hp ^ 2) = 2 (MH ^ 2 + ho ^ 2) = 2om ^ 2, so, (PM ^ 2 + PN ^ 2) / AB ^ 2 = 2om ^ 2 / (2om) ^ 2 = 1 / 2. It can be seen that when P changes, the ratio remains unchanged, which is always 1 / 2