If point P is Fermat point of acute triangle ABC, and angle ABC = 60 degrees, PA = 3, PC = 4, then the value of Pb is________ ； (I'm a sophomore) don't go beyond your ability

If point P is Fermat point of acute triangle ABC, and angle ABC = 60 degrees, PA = 3, PC = 4, then the value of Pb is________ ； (I'm a sophomore) don't go beyond your ability

According to the definition of Fermat point, APB, BPC and CPA are all 120 degrees. Below, take BC as an edge and make an equilateral triangle △ BCM outward. Then, according to the proof process of Fermat point, P takes point n on am (this step can be Baidu Encyclopedia) and PM, so that ∠ PBN = 60 degrees, from ∠ APB = 120 degrees, so ∠ BPN = 60 degrees, so △ BPN is positive three

Proof: BB 'passes through the Fermat point P of △ ABC, and BB' = PA + Pb, why are four points a, B ', P and C in the proof co circular?
If P is a point on the plane where △ ABC is located, and ∠ APB = ∠ BPC = ∠ CPA = 120, then p is called the Fermat point of △ ABC. On the outside of the acute angle △ ABC, make an equilateral △ ACB 'to connect BB' to prove that BB 'passes through the Fermat point P of △ ABC, and BB' = PA + Pb + PC
It is proved that: from ∠ BPA = 120 °, ab ′ C = 60 °,
The four points a, P, C and B 'are in the same circle
∴∠APB′=∠ACB′=60°,
∴∠APB+∠APB′=180°,
BPb ′ three points are collinear
Take a point D on PB 'so that ∠ PCD = 60 °,
From ∠ CPB ′ = 120 ° - 60 ° = 60 °,
The results show that PC = PD (1),
In △ APC and △ B ′ DC,
AC = B ′ C, from ∠ PCD = ∠ ACB ′ = 60 °,
∴∠ACP=∠B′CD,PC=DC,
≌△ ACP ≌△ B ′ CD, AP = DB ′ (2)
From (1) and (2), it is concluded that:
BP + AP + CP = BB '

B 'is too much trouble. Use d instead
Make an equilateral triangle ACD outside the acute triangle ABC to connect BD and CD
Because it is an acute triangle, you can select a point P on BD to make the angle APB equal to 120 degrees
Then the angle APD equals 60 degrees and the angle ACD equals 60 degrees
So the four points a, P, C and D are all round
So angle cpd equals angle CAD equals 60 degrees
In conclusion, the angle APC is equal to the angle BPC is equal to 120 degrees, so p is the Fermat point
Xiazheng BD = PA + Pb + PC
Extend PC and intercept CE = AP, connect ed
Because of the internal connection, it is easy to know that the angle Pad + angle PCD = 180 degrees, and the angle pad = angle ECD
The easy syndrome is △ pad ≌ △ ECD, and ∵ angle dep = angle DPE = 60 degrees
∴PD=ED=PE=PC+CE=PC+PA
It is proved that BD = PA + Pb + PC

Make an equilateral triangle ACB "on the outside of the acute triangle ABC and connect BB" to prove that BB "passes through the Fermat point P of the triangle ABC and BB" = PA + Pb + PC

It is proved that: (1) if we take point P on BB 'such that the angle APC = 120 degrees, then PABC is four points co circle, then the angle APB = ACB = 60 degrees, CPB = cab = 60 degrees, so the angle APB = CPB = 120 degrees, then p is a Fermat point. (2) if we take B'd = PA, then PA = db', AC = B'c, PAC = DBC, then the triangle APC is all equal to BDC, so PC = CD and BPC = 60 degrees, then the triangle PCD is equal

As shown in the figure, the radii of two concentric circles with o as the center are 5 and 3 respectively. If the chord ab of the big circle intersects the small circle at points c and D, the value range of chord AB is______ ．

When AB is tangent to the small circle, ∵ the radius of the big circle is 5cm, the radius of the small circle is 3cm, ∵ AB = 252 − 32 = 8cm

As shown in the figure, in the two concentric circles with o as the center, the chord ab of the big circle intersects the small circle C and D
If the lengths of AC and AD are exactly the two real roots of the equation x & sup2; - MX + 6 = 0, can we find out the area of the ring inside the big circle and outside the small circle? If we can, ask for its area. If we can't, what conditions do you think we need to add to find out its area

You can ask
S=π（R²-r²）=6π

As shown in the figure, in two concentric circles with o as the center, the chord ab of the big circle intersects the small circle at two points c and D

For OE ⊥ AB, then AE = be, CE = De, so be-de = ae-ce; that is, AC = BD

As shown in the figure, we know two concentric circles with point o as the common center. The chord ab of the big circle intersects the small circle with C, D. (1) prove: AC = db; (2) if AB = 6cm, CD = 4cm, find the area of the ring

(1) Through the point o as OE ⊥ AB in E, ≁ AE = be, CE = De, ≁ ae-ce = be-de, ≁ AC = BD; (2) connect OA, OC, in RT △ AOE and RT △ OCE: oe2 = oa2-ae2, oe2 = oc2-ce2, ≁ oa2-ae2 = oc2-ce2, ∵ oa2-oc2 = ae2-ce2, ∵ AB = 6cm, CD = 4cm, ≁ AE = 3cm, CE = 2cm

As shown in the figure, we know two concentric circles with point o as the common center. The chord ab of the big circle intersects the small circle with C, D. (1) prove: AC = db; (2) if AB = 6cm, CD = 4cm, find the area of the ring

(1) Through the point o as OE ⊥ AB in E, ≁ AE = be, CE = De, ≁ ae-ce = be-de, ≁ AC = BD; (2) connect OA, OC, in RT △ AOE and RT △ OCE: oe2 = oa2-ae2, oe2 = oc2-ce2, ≁ oa2-ae2 = oc2-ce2, ∵ oa2-oc2 = ae2-ce2, ∵ AB = 6cm, CD = 4cm, ≁ AE = 3cm, CE = 2cm

As shown in the figure, in the concentric circle O, the chords AB and AC of the big circle are cut into smaller circles D and e respectively

Connect OD, OE, OA, then od = OE; OA = OA, ∠ ODA = ∠ OEA = 90 degrees, so Δ AOD ≌ Δ AOE; so ad = AE. Because AB = 2ad, AC = 2ae; so AB = AC, that is: ∠ B = ∠ C

As shown in the figure: in a concentric circle, the chord ab of the big circle intersects the small circles C and D. It is known that ab = 2CD, and the chord center distance of AB is equal to half of CD. What is the radius ratio of the big circle and the small circle of the concentric circle?

Let OE ⊥ CD be set at point E, and OE = 1
∵OE=1/2CD
∴CE =CE=1
∴OC =√2
∵AB =2CD
∴AC =CE=1
∴AE=2
∴OA=√5
The radius ratio of small circle to large circle is √ 2 ∶ 5