Straight line and plane vertical exercises The triangle ABC has a point p outside the plane α, through which PQ is made, the vertical plane α is made, and the perpendicular foot is O, connecting PA, Pb and PC (1) If PA = Pb = PC, then o is triangular ABC___ Heart (2) If PA vertical Pb, Pb vertical PC, PC vertical PA, then o is triangular ABC___ Heart (3) If the distances from P to three sides AB, BC and Ca are equal, then o is a triangle ABC___ Heart (4) If PA = Pb = PC, ∠ C = 90 °, then o is ab edge___ . (5) If PA = Pb = PC, ab = AC, then point O is at___ on-line. There is no picture in the original question. Yes, please send me the answer and the answer

Straight line and plane vertical exercises The triangle ABC has a point p outside the plane α, through which PQ is made, the vertical plane α is made, and the perpendicular foot is O, connecting PA, Pb and PC (1) If PA = Pb = PC, then o is triangular ABC___ Heart (2) If PA vertical Pb, Pb vertical PC, PC vertical PA, then o is triangular ABC___ Heart (3) If the distances from P to three sides AB, BC and Ca are equal, then o is a triangle ABC___ Heart (4) If PA = Pb = PC, ∠ C = 90 °, then o is ab edge___ . (5) If PA = Pb = PC, ab = AC, then point O is at___ on-line. There is no picture in the original question. Yes, please send me the answer and the answer

(1) PA = Pb = PC, that is OA = ob = OC, then o is a triangle ABC_ Outside__ Center [circumscribed circle center]
(2) Because PA ⊥ Pb, Pb ⊥ PC, so Pb ⊥ face PAC, so Pb ⊥ AC, and Po ⊥ AC, so
AC ⊥ face PbO, so ob ⊥ AC
Similarly, OA ⊥ BC, OC ⊥ ab
O is triangular ABC_ Hang__ Heart [the intersection of three heights]
(3) That is to say, the distance from O to the three sides is equal, and O is a triangle ABC_ Inside__ Center [center of inscribed circle]
(4) (1) it is mentioned that O is the outer center, that is, the center of circumscribed circle, and ∠ C = 90 °, so AB is the diameter, so o is the diameter of AB side__ Midpoint__
(5) If AB = AC, then o is on the vertical line of BC
[mathematics, physics and chemistry in middle school]

How to determine the angle between a straight line and a plane and the angle between a plane and a plane is a problem in solid geometry

Build a system and solve it by vector

The problem or method of dihedral angle evaluation in solid geometry is no less than 10 questions

You can search in Baidu "2013 high school mathematics solid geometry dihedral problem", the first Baidu Library is. Welcome to ask, hope to adopt!

As shown in the figure, AB is the diameter of ⊙ o, and the chord CD ⊥ AB is at e, and the tangent line passing through point B intersects the extension line of ad at point F. (1) if M is the midpoint of AD, connect me and extend me to intersect BC at n. verification: Mn ⊥ BC. (2) if cos ∠ C = 45, DF = 3, calculate the radius of ⊙ o

(1) It is proved that: (1) connecting AC. ∵ AB is the diameter of ⊙ o, and ab ⊥ CD is in E. according to the vertical diameter theorem, point E is the middle point of CD; and ∵ m is the middle point of AD, ∵ me is the median line of △ DAC, ∵ Mn ∥ AC. ∵ AB is the diameter of ⊙ o, ∵ ACB = 90 °, that is Mn ⊥ BC

As shown in the figure, AB is the diameter of the circle O, CD is the tangent line of the circle O, C is the tangent point ad, the extension line perpendicular to CD ad intersects with the extension line BC at the point e to prove AE = ab

(1) When AC is the bisector of ∠ bad, ad ⊥ CD,
Proof: connect BC,
Then ∠ ACB = 90 degree, that is ∠ B + ∠ BAC = 90 degree,
∵ CD is the tangent of circle o,
∴∠ACD=∠B,
∵ AC bisection ∠ bad,
∴∠BAC=∠CAD,
∴∠CAD+∠ACD=90°,
That is, d = 90 ° ad ⊥ CD;
(2) It can be seen from (1): BAC = CAD,
∵∠ACB=∠D=90°,
∴△ABC∽△ACD,
∴AD：AC=AC：AB,
∴AC2=AD•AB=20；
The solution is AC = 2,
In right triangle ACD,
According to Pythagorean theorem, CD = 2,
According to the fact that CD is the tangent of a circle, we can get: CD2 = ad &; De, that is, de = CD2 ﹣ ad = 4 ﹣ 4 = 1. So AB = AE

As shown in the figure, AB is the diameter of ⊙ o, CD is the tangent of ⊙ o, and the tangent point is C. extend AB, intersect CD at point E. connect AC, make ∠ DAC = ∠ ACD, AF ⊥ ed at point F, intersect ⊙ o at point g. (1) prove that ad is the tangent of ⊙ o; (2) if the radius of ⊙ o is 6cm, EC = 8cm, find the length of GF

(1) It is proved that: to connect OC, CD is the tangent of \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\it's not easy= 90 °, af6 = 1610. AF = 9.6. ∵ AB is the diameter of ⊙ o, ∵ AGB = 90 °. ∵ AGB = ∵ AFE. ∵ bag = ∵ EAF, ∵ RT ∵ ABG ∵ RT ∵ AEF. ∵ agaf = abae. Ag9.6 = 1216. ∵ Ag = 7.2. ∵ GF = af-ag = 9.6-7.2 = 2.4 (CM)

As shown in the figure, AC ⊥ BC is at point C, BC = a, CA = B, ab = C, ⊙ o is tangent to the straight line AB, BC, CA, then the radius of ⊙ o is equal to______ ．

Let the tangent points of AC, Ba, BC and ⊙ o be D, F and e respectively; connect od and OE; ∵ AC and be are tangent lines of ⊙ o, ∵ ODC = ∵ OEC = ∵ DCE = 90 °; ∵ quadrilateral odce is rectangle; ∵ od = OE, ∵ rectangular odce is square; namely OE = od = CD; let CD = CE = x, then ad = AF = b-X; connect ob, of, by Pythagorean

Point D is a point on the extension line of the diameter CA of the center O, point B is on the center O, and ab = ad = Ao
(1) Verification: BD is the tangent of ⊙ o;

Certificate: connect Bo
∵ AB = Ao = Bo, that is, △ ABO is an equilateral triangle
∴∠BAO=∠ABO=60°
∵AD=AB,∠BAO=60°
∴∠ABD=∠ADB=30°
∴∠DBO==∠ADB+∠ABO=90°
That is BD ⊥ Bo
⊙ BD is the tangent of ⊙ o

As shown in the figure, the trapezoid ABCD is connected to the circle O, ad is parallel to BC, the tangent line passing through the leading circle O of B intersects Da respectively, and the extension line of Ca intersects E and F
《1》 Prove the square of AB = AE * BC
Given BC = 8, CD = 5, AF = 6, find the length of EF

The first problem is that ∵ BF cuts ⊙ o to B, ∵ Abe = ∵ BCA. ∵ ad ∥ BC, ∵ EA ∥ BC, ∵ BAE = ∵ ABC. From ∵ Abe = ∵ BCA, ∵ AE / AB = AB / BC, ∵ AB ^ 2 = AE × BC

As shown in the figure, the rectangle ABCD is inscribed in circle O, ad is parallel to BC, and the tangents passing through the leading circle O of B intersect the extension lines of Da and Ca at e and f respectively
1 prove the square of AB = AE times BC
Given that BC equals 8, CD equals 5, AF equals 6, find the length of EF