Seeking high school solid geometry problem It's a little difficult to find some intersection lines in high school solid geometry. It's also a proof of solid geometry

Seeking high school solid geometry problem It's a little difficult to find some intersection lines in high school solid geometry. It's also a proof of solid geometry

As shown in the figure, in the pyramid p-abcd, PD ⊥ plane ABCD, ad ⊥ CD, DB bisection ∠ ADC, e is the midpoint of PC, ad = CD = 1, & nbsp; (1) prove PA ∥ plane BDE; (2) prove AC ⊥ plane PBD; (3) find the tangent of the angle between the line BC and plane PBD

On the three perpendicular theorem in solid geometry
Can the three perpendicular theorem be applied without forming a triangle? That is to say, the line on the bottom surface is not at the projection position of the oblique line?
If you know, please say,

In the three perpendicular theorem, three lines mean an oblique line L with an intersection point with a plane, the projection L 'on the inclined plane and the straight line a perpendicular to the projection on the plane,
There are l '⊥ a ← → L ⊥ a
Dihedral angle solution is how to solve the problem of two disjoint lines, the main problem is the auxiliary line, how to do, how to do
Find a face in the graph that is parallel to one of the two lines. Make sure that the face has an intersection with the other line
Then we need to do the analysis of the angle of the triangle, not to mention the special contour, the rest need to do the high line, use the trigonometric function to find the angle, and refer to Sina / cosa / Tana / COTA

As shown in the figure, the planes of two congruent squares ABCD and abef intersect AB, m ∈ AC, n ∈ FB, and am = FN

Proof 1: MP ⊥ BC, NQ ⊥ be through M, P and Q are perpendicular feet (as shown in the figure), connecting PQ. ∵ MP ∥ AB, NQ ∥ AB, ∥ MP ∥ NQ. NQ = 22bn = 22cm = MP, ∥ mpqn is parallelogram. ∥ Mn ∥ PQ, PQ ⊂ plane BCE. And Mn ⊄ plane BCE, ∥ Mn ⊉ plane BCE. Proof 2: Mg ∥ BC through M, intersect

As shown in the figure, PA tangent ⊙ o at point a, the radius of the circle is 3, Po = 5, then the length of PA is equal to______ ．

∵ PA ⊙ o is tangent to point a, ∵ OA ⊥ AP; in RT △ AOP, OA = 3, Po = 5; according to Pythagorean theorem, PA = op2 − oa2 = 4

If PA = 1, ab = 2, Po = 3, then the radius of circle O is equal to
If OM is perpendicular to point m, then BM = am = 2 △ 1 = 1
MP=1+1=2
OP=3
In RT △ OPM, the Pythagorean theorem om = radical (3 & # 178; - 2 & # 178;) = radical 5
A: the radius is 5
What's wrong with the above problem solving process?
Mark the intersection of Po and circle as C, extend the intersection of Po and circle D, and connect AD and BC
∵ ∠B=∠D ∠P=∠P
∴ △APD ∽△CPB
AP/CP = DP/BP
∵ BP=AP+AB=1+2=3 CP=OP-OC DP=OP+OD
OC = od is equal to the radius of ⊙ o, expressed by R
∴ 1 / (3-r) = (r+3)/(1+2)
r = √6
The radius of circle O is equal to √ 6
Is the solution right_ ⊙)？ Why do I count in the top one?

The mistake: you mistook om for radius,
Next, we need to continue to further link OA
In the OAM of right triangle, by Pythagorean theorem, we get the following results
Radius OA = root (OM ^ 2 + am ^ 2)
=Root (1 ^ 2 + 5)
=Root 6
Your answer below is correct, but it's too complicated

As shown in Fig. 1 + two strings PA and Pb are made at a point P over circle O. if PA = Pb, then Po bisects ∠ APB. Why?

If the chords are equal, then the distance between the chords and the center of the chords is equal, and ﹥ Po bisects ﹥ APB (the equal points on both sides of the angle are on the bisector of the angle)

Given that P is a point on the circle O, two strings PA and Pb are drawn from P, and Po bisection ∠ APB is obtained

If PA = Pb, then Po bisects ∠ APB
Connect OA, ob
∵ PA = Pb, Op = OP, OA = ob (radius)
∴△AOP≌△BOP
∴∠APO=∠BPO
Ψ OP bisection ∠ APB

The radius of circle 0 is 1, P is a point outside circle 0, PA is tangent to circle 0 at point a, PA = 1, AB is the chord of circle 0, and ab = two root sign 2, find Pb

Connect ob, OA = ob = 1, ab = radical 2
So △ AOB is an isosceles right triangle
Because AP tangent circle o
So OA ⊥ pa
Because ∠ OAB = 45 degrees
Therefore, PAB = 45 degrees
Because ∠ OAB = ∠ PAB, AP = Ao, ab = ab
So △ PAB ≌ △ OAB (SAS)
So Pb = ob = 1

If the circle O with PA tangent radius 1 is at point a, PA = 1 and chord AB = root 2, then Pb =?

In the triangle ABO, 2 ^ 2 + 2 ^ = (2 root 2) ^ 2, that is, OA ^ 2 + ob ^ 2 = AB ^ 2 and OA = ob, so the triangle OAB is isosceles right triangle, so the angle OAB = 45 degrees and PA is tangent, and the angle Pao = 90 degrees, so the angle PAB = angle Bao = 45 degrees and ab = AB, Ao = AP = 2 triangle

Given that the radius of the circle O is 1, P is a point outside the circle O, PA is tangent to the circle O at the point a, PA = 1, AB is the chord of the circle O, and ab = √ 3, then the value of the square of Pb is to find the great God

You say that ab =? 3, come on! I'm counting!