It is known that P is a point in the plane of triangle ABC. If vector CB = γ vector PA + vector Pb, γ belongs to R, then point P must be in the plane A. Inside the triangle ABC B. On the line where the AC side is located C. On the line where AB side is located D. On the line where the BC side is located Why?

It is known that P is a point in the plane of triangle ABC. If vector CB = γ vector PA + vector Pb, γ belongs to R, then point P must be in the plane A. Inside the triangle ABC B. On the line where the AC side is located C. On the line where AB side is located D. On the line where the BC side is located Why?

Vector CB = γ vector PA + vector Pb, γ belongs to R
CB + BP = YPA, that is CP = YPA. A, C, P are collinear

In the pyramid p-abcd, the plane pad ⊥ the plane ABCD. ∠ ABC = ∠ BCD = 90 °, PA = PD = DC = CB = 1 / 2Ab, e is the midpoint of Pb Conclusion: (1) EC is parallel to plane APD (2) Find the tangent of the angle formed by BP and plane ABCD

Proof: (1)
Take point m of AB and connect cm and em
In △ BPA, me is the median line, ν me ‖ PA
In the quadrilateral ABCD,
∵∠ABC=∠BCD=90°,DC=1/2AB=AM
The quadrilateral ADCM is a parallelogram (BC is parallel and equal to AM)
Then MC ∥ ad
/ / face CEM ∥ plane APD (a pair of intersecting lines are parallel)
Then CE ‖ surface APD
（2）
Take the midpoint n of AD and connect PN and BN
∵PA=PD
∴PN⊥AD
∵ plane pad ⊥ plane ABCD
⊥ plane ABCD
Then PN ⊥ BN ∠ PBN is the angle formed by BP and ABCD
Connecting DM and BD
∵DC∥MB,BC=DC=AB/2=MB,∠ABC=∠BCD=90°
The quadrilateral BCDM is a square
Let AB = 2A
Then am = MB = BC = EC = DM = PA = Pb = a
AD=MC=DB=√2a
DN=AD/2=√2/2 a
PN²=PD²-DN²=a²-1/2a²=1/2a²,PN=√2/2a
∵ DM⊥AB,MD=MB=MA
Ψ MDA = ∠ MDB = 45 °, i.e. ∠ BDA = 90 °
BN²=BD²+DN²=2a²+1/2 a²=5/2a² BN=√10/2a
tan∠PBN=PN/BN= √2/2a / √10/2a =√5/5

As shown in the figure, PA, Pb are the tangent lines of ⊙ o, and the tangent points are a, B, and the straight line EF is also the tangent line of ⊙ o, the tangent point is Q, EF intersects PA respectively, Pb is at e, f points, PA is known to be equal to α, ∠ P = α, (1) calculate the circumference of ⊙ PEF; (2) calculate

According to the title:
EA=EQ,FB=FQ,PA=PB=10
∴C△PEF=PE+PF+EF=PE+PF+EQ+FQ=PE+PF+EA+FB=PA+PB=20
Connect Ao, Qo, Bo
Easy to get: △ AOE ≌ △ QoE, △ BOF ≌ △ qof
∴∠AOE=∠QOE,∠BOF=∠QOF
∵∠P=70°
∴∠AOE+∠QOE+∠BOF+∠QOF=110°
∴∠EOF=∠EOQ+∠FOQ=55°

PA, Pb are tangent lines of circle O, tangent EF tangent circle O to C and intersection PA to E. if PA = 6cm, the circumference of triangle PEF is

∵ PA, Pb are tangent lines of circle o
∴PA=PB=6
∵ tangent EF tangent circle O to C, PA to e, Pb to F,
∴BF=CF
AE=CE
The circumference of △ PEF
=PF+EF+PE
=PF+CF+CE+PE
=PF+BF+AE+PE
=PB+PA
=6+6
=12

PA, Pb are tangent lines of circle O, EF intersects PA, Pb at point E and F, tangent point C is on circle O. if the circumference of △ PEF is 4, what is the length of PA Now it's still out of date

According to the tangent length theorem
PA=PB,EA=EC,FB=FC
Δ PEF perimeter = PE + pf + ef
=PE+PF+EC+FC
=PE+PF+EA+EB
=PA+PB
=2PA
So 2PA = 4
PA=2

PA, Pb and circle O are tangent to point a and B respectively. Tangent EF of circle O intersects PA, Pb at point E and F, tangent point C is on arc ab. if PA length is 2, then perimeter of triangle PEF is

EA=EC,FB=FC,PA=PB=2
C△=PE+PF+EF=PE+PF+EC+FC=PE+PF+EA+FB=PA+PB=4

As shown in the figure, P is a point outside ⊙ o, PA, Pb and ⊙ o are tangent to points a and B respectively, and C is a point outside ⊙ o At any point on AB, the tangent passing through point C intersects PA and Pb at point D and e respectively (1) If PA = 4, find the circumference of △ ped; (2) If ∠ P = 40 °, find the degree of ∠ doe

(1) ∵ Da, DC are tangent lines of circle o,
∴DC=DA，
Similarly, EC = EB, PA = Pb,
The circumference of triangle PDE = PD + PE + de = PD + DC + PE + be = PA + Pb = 2PA = 8,
The perimeter of triangle PDE is 8;
（2）∵∠P=40°，
∴∠PDE+∠PED=140°，
∴∠ADC+∠BEC=（180-∠PDE）+（180-∠PED）=360°-140°=220°，
∵ Da, DC are tangent lines of circle o,
∴∠ODC=∠ODA=1
2∠ADC；
In the same way: ∠ OEC = 1
2∠BEC，
∴∠ODC+∠OEC=1
2（∠ADC+∠BEC）=110°，
∴∠DOE=180-（∠ODC+∠OEC）=70°．

AC is the diameter of circle 0, AC = 10cm, PA and Pb are tangent lines of circle 0, a and B are tangent points, and ad vertical BP is made through a, and the intersection of BP with BP is made

Connect OP and AB to point E
∵ PA, Pb are tangent lines of ⊙ o
ν Po vertically bisects ab
∵ PA is the tangent of ⊙ o,
∴OA⊥PA
∵PA＝12,OA＝5
According to Pythagorean theorem, Op = 13
Using the area of triangle, we can get: PA × Ao = Po × AE
∴AE＝60/13
∴AB =120/13cm

PA, Pb are two tangent lines of ⊙ o, a and B are tangent points, secant PCD intersects ⊙ o at C and D, and connects a, C, B, D in order. Verification: AC · BD = ad · BC On the second floor. Why is the angle PBC equal to angle PDB and angle DPB equal to angle DPB in triangle PDB and triangle PBC?

Because in triangle PDB and triangle PBC, the angle PBC is equal to angle PDB, and angle DPB is equal to angle DPB
So Pb / PD = BC / BD
In the same way
PA/PD=AC/AD
Because the length of the circle is the same
So PA = Pb
So BC / BD = AC / ad
Ac * BD = ad * BC

AC is the diameter of circle O, AC = 10 Pb, PA is the circle O tangent line ad is perpendicular to BP, connect AB, BC, AP = 12, and calculate the length of ab

Connecting opoa = 5, AP = 12, with the help of Pythagorean theorem, we can get the triangle OP = 13 and OAP is a right triangle. It can be proved that OP is perpendicular to AB, and if the perpendicular foot is e, then AE is the height on the oblique side of the right triangle OAP, which can be obtained and multiplied by 2