As shown in the figure, AB is perpendicular to plane BCD, BC is perpendicular to CD, ab = BC, and the angle between AD and plane BCD is 30 degrees Finding the angle between AD and plane ABC Picture in

Because ab ⊥ plane BCD, the line CD is in plane BCD
So ab ⊥ CD and ∠ DAB is the angle between AD and plane BCD
Then ∠ DAB = 30 °
Also BC ⊥ CD, and AB.BC Is two intersecting lines in plane ABC
Therefore, it can be seen from the determination theorem of the perpendicularity of lines and planes
CD ⊥ plane ABC
Then, DAC is the angle between AD and plane ABC
In RT △ ABC, ab = BC
Then AC = √ 2 * AB can be obtained from Pythagorean theorem
In RT △ abd, DAB = 30 degree
Then ad = 2Ab
Therefore, in RT △ ACD, ∠ ACD = 90 degree
There are cos ∠ DAC = AC / ad = √ 2 * AB / (2Ab) = √ 2 / 2
Then ∠ DAC = 45 °
So the angle between AD and plane ABC is 45 degrees

It is known that in △ BCD, ∠ BCD = 90 °, BC = CD = 1, ab ⊥ plane BCD, ∠ ADB = 60 °, e and F are points on AC and ad respectively, and AE: AC = AF: ad = λ (0
(1) Proof: no matter what the value of λ is, there is always plane bef ⊥ plane ABC
(2) When λ is, plane bef ⊥ plane ACD

1. Because ab ⊥ plane BCD, so AB is vertical to CD, and because BC ⊥ CD, so CD ⊥ plane ABC, because AE: AC = AF: AD, so EF is parallel to CD, so EF ⊥ plane ABC, so plane bef ⊥ plane ABC2, because EF ⊥ plane ABC, so EF ⊥ be, so be ⊥ AC is needed when plane bef ⊥ ACD, at this time, BD = root 2

It is known that ab ⊥ plane BCD, m and N are the midpoint of AC and ad respectively, BC ⊥ CD. It is proved that (1) Mn ⊥ plane BCD; (2) plane BCD ⊥ plane ABC

Certification:
1. In △ ACD, because Mn is the midpoint of AC and ad respectively, there is Mn ‖ CD
But CD is in plane BCD
So: Mn ‖ BCD
2. Because ab ⊥ is BCD, and ab is in plane ABC
So: plane ABC ⊥ plane BCD

The intersection of BP and AQ of any triangle ABC PQ on AC and BC, P and Q on BC and AC is o. if the areas of BOQ, ABO and apo are 123, find s PQC

Let S1 = s ⊿ CoQ, S2 = s ⊿ cop, S3 = s ⊿ BOQ, S4 = s ⊿ AOB, 5 = s ⊿ AOP. S4: S3 = (S5 + S2): S1 2S1 = 3 + S2 (1) S5: S2 = S4: (S3 + S1) 3 + 3s1 = 2s2 (2) S1 = 9, S2 = 15s ⊿ ABC = S1 + S2 + S3 + S3 + S5 = 30ap: PC = S5: S2 = 3:15 = 1:5pc: AC = PC: (PC + AP) = 5:6bq: QC = S3: S1

In the triangle ABC, ab = 5, BC = 3, AC = 4, PQ is parallel to point AB, P is on AC, and point q is on BC if the triangle PQC
The area of is equal to the area of quadrilateral pabq to find CP

In △ ABC, ab = 5, BC = 3, AC = 4
So △ ABC is a right triangle and ab is a hypotenuse
Area of △ ABC = 3 * 4 / 2 = 6
When the area of △ PQC is equal to the area of quadrilateral pabq
So △ PQC = 3
PQ‖AB
CP：4=CQ:3
CQ=3CP/4
△PQC=1/2*CQ*CP=3/8*CP^2=3
CP = 2 * radical 2

D is a point on the triangle ABCA, e is a point in the triangle ABC, de ∥ BC, make a parallel line crossing CE through D, extend the line to F, CF and ab intersect at point P, prove that BF is parallel to Ae

Proof: due to de / / BC
So Pd / Pb = PE / PC
Due to FD / / AC
So PA / PD = PC / PF
So PA / Pb = PE / PF
Because angle ape = angle BPF
So triangle ape is similar to triangle BPF
So angle FBP = angle EAP
So BF / / AE

As shown in figures (1) and (2), Mn is the diameter of ⊙ o, the chord AB and CD intersect at a point P on Mn, ∠ APM = ∠ CPM
(1) Based on the above conditions, what do you think is the size relationship between AB and CD? Please explain the reasons. (2) if the intersection P is outside ⊙ o, is the above conclusion true? If yes, prove it; if not, give reasons

(1) AB = CD, the reason is: o as OE ⊥ AB in E, of ⊥ CD in F, connect ob, OD, ∵ ⊥ APM = ⊥ CPM, ∵ APM = ⊥ BPN, ∵ CPM = ⊥ DPN, ∵ BPN = ⊥ DPN, ∵ OE ⊥ AB, of ⊥ CD, ≁ OE = of, in RT △ BeO and RT △ DOF, of = OE, OD = ob, obtained by Pythagorean theorem: be = DF

Mn is the diameter of circle O, the chord AB and CD intersect at a point P on Mn, and PD = Pb. To prove: ab = CD, we need to give a detailed solution

Proof: connect ob, OD
For OE ⊥ CD in E, of ⊥ AB in F
∵ in circle O, OD and ob are radius
∴OD=OB
∵OD=OB
PD=PB
PO=PO
∴△PDO＝△PBO（SSS）
∵∠ DPN = ∠ BPN  PN bisection ∠ DPB
∵OE⊥CD,OF⊥AB
∴OE=OF
In the circle O, ab = CD

AB is the diameter of circle O, and ab = 10, if both ends of chord Mn slide on the circumference
AB is the diameter of circle O, and ab = 10. If the length of chord Mn is 8, the two ends of Mn slide on the circumference and always intersect ab. the distance between counting points a and B and Mn is H1, H2, and / h1-h2 / (/ / represents the absolute value)

Let Mn and ab be perpendicular to each other, let them intersect at point P, then AP is the distance from a to Mn, BP is the distance from B to Mn, obviously MP = NP = 4, so OP = 3, Pb = 2, so | h1-h2 | = (3 + 5) - 2 = 6
In another special case, if a point in Mn coincides with a or B, let n coincide with B. then, because AB is the diameter, am is the distance from a to Mn, am = sqrt (10 × 10-8 × 8) = 6, and the distance from B to Mn is 0, so the answer is 6
For general cases, let AB cross Mn to P, let o cross the center of circle as the vertical line of Mn to C, let B cross the vertical line to e, let a cross the vertical line to D, let AP = x, then OP = 5-x, OB = 10-x, it is easy to know that OC = 3, there is a triangle APD similar to OPC, AP / op = ad / OC, so ad = 3x / (5-x), and the triangle POC is similar to PBE, OC / be = op / BP, so be = 3 × (10-x) / (5-x), so be-ad = [3 (10-x) - 3x] / (5-x) = 6

As shown in the figure, AB is the diameter of the circle O, and ab = 10. The length of the chord Mn is 8. If the two ends of the chord Mn slide on the circle, they always intersect with ab. note the distance from a, B to Mn
H 1 and H 2, then what is the absolute value of H 1 - H 2? Draw your own picture

H1 + H2 = two times the distance from the center O to Mn. Using the vertical diameter theorem, we get that if the distance is 3, then H1 + H2 = 6

As shown in the figure, AB is perpendicular to plane BCD, BC is perpendicular to CD, ab = BC, and the angle between AD and plane BCD is 30 degrees Finding the angle between AD and plane ABC Picture in