It is known that, as shown in the figure, △ ABC is an inscribed regular triangle of ⊙ o, and point P is a moving point on arc BC. This paper explores the relationship between PA, Pb and PC

It is known that, as shown in the figure, △ ABC is an inscribed regular triangle of ⊙ o, and point P is a moving point on arc BC. This paper explores the relationship between PA, Pb and PC

PA=PB+PC.
Reason:
Intercept PD = Pb on PA and connect BD,
∵ Δ ABC is an equilateral triangle,  AB = BC,  ABC = ∠ C = 60 °,
Ψ P = ∠ C = 60 °, Δ PBD is an equilateral triangle,
∴PB=BD,∠PBD=∠PDB=60°,
Ψ ABC - ∠ CBD = ∠ PBD - ∠ CBD, that is ∠ abd = ∠ CBP,
In Δ abd and Δ CBP:
AB=BC,∠ABD=∠CBP,BD=PB,
∴ΔABD≌ΔCBP(SAS),
∴AD=PC,
∴PA=PB+PC.

If p 'is a point outside △ ABC and △ p'ab ≌ △ PAC, find the distance between points P and P' and the degree of ∠ APB

∵△P’AB≌△PAC
∴∠P’AB=∠PAC
∵∠BAP+∠PAC=60°
∴∠P'AB+∠BAP=60°
∵P'A=PA,∠P'AP=60°
Connect p'p
ν Δ p'ap is an equilateral △
∵P'A=PA=6
∴P'P=PA=6
∵P'B=PC=10,P'P=6,BP=8
Δ p 'is a right angle
∴∠APB=90°+60°=150°

As shown in the figure, P is any point in the equilateral △ ABC, connecting PA, Pb and PC （1）PA+PB+PC＞3 2AB； （2）AP+BP＞PC． (Note: only triangle trilateral relationship is used to prove)

(1) ∵ PA + Pb > AB Pb + PC > BC PC + PA > AC, ∵ AB + Pb + Pb + PC + PC + PA)

As shown in the figure, P is a point in the equilateral triangle ABC, and PA = 6, Pb = 8, PC = 10. If the triangle PAC is rotated anticlockwise around point a, the triangle p'ab is obtained, and the distance between point P and point P 'and the size of angle APB are calculated

Suppose that the length of the edge of the equilateral △ ABC is k, then BD = K / 2
AD²
=AB²-BD²
=k²-k²/4
=3k²/4
AD=(√3)k/2
Area s = 1 / 2 × BC × ad = 1 / 2 × K × (√ 3) K / 2 = (√ 3) K / 4
Take PA as the edge direction △ ABC, make an equilateral △ ape (e point is outside AB side), and connect be, we can know: ∠ BAE + ∠ PAB = ∠ BAC = ∠ PAE = ∠ cap + ∠ PAB = 60 ° so: ∠ BAE = ∠ cap; ab = AC, AE = AP, therefore, △ BAE ≌△ cap; then: be = CP = 10,
In △ BPE, PE = 6, Pb = 8, be = 10, because: 6? + 8? = 10
Therefore, △ BPE is a right triangle with ∠ BPE as its right angle, so: ∠ APB = ∠ APE + ∠ BPE = 60 ° + 90 ° = 150 °,
In △ ABP, the cosine theorem is used to obtain the following results
k²=AB²=PA²+PB²-2×PA×PB×cos∠APB
=6²+8²-2×6×8×cos150°
=100+48√3
To sum up, s △ ABC = (√ 3) K 2 / 4 = (√ 3) / 4 × (100 + 48 √ 3) = 25 (√ 3) + 36
Note: in the computer, √ means the second root, 25 (√ 3) + 36 means 25 times the root 3, plus 36

As shown in the figure, △ ABC, the vertical bisectors of edges AB and BC intersect at point P (1) Confirmation: PA = Pb = PC; (2) Is point P also on the vertical bisector of edge AC? What else can you draw from that?

It is proved that: (1) the vertical bisectors of edges AB and BC intersect point P,
∴PA=PB，PB=PC．
∴PA=PB=PC．
（2）∵PA=PC，
The point P is on the vertical bisector of edge AC (the point with the same distance from the two endpoints of a line segment is on the vertical bisector of the line segment)
It can also be concluded that: (1) the vertical bisectors of the three sides of the triangle intersect at a point
② This point is equidistant from the three vertices

As shown in the figure, in the triangle ABC, ad is the bisector of ∠ BAC, EF bisects ad vertically, intersects ad at e, and the extension line intersects BC at F, then ∠ B = ∠ caf

Because ad bisects angle BAC, angle bad = angle CAD
Because EF bisects ad vertically, AF = DF, so angle DAF = angle ADF
Because angle DAF = angle DAC + angle CAF, angle ADF = angle B + angle bad, angle DAC + angle CAF = angle B + angle bad
Because angle bad = angle CAD, angle B = angle caf

As shown in the figure, in the triangle ABC, the ad bisection angles BAC, E.F. are BD.AD And de = CD, EF = AC stay

It is proved that if FD is extended to m, DM = DF, and de = CD, then ⊿ CDM ≌ ⊿ EDF (SAS), ∠ EFD = ∠ CMD; cm = EF
If EF = AC, then cm = AC, ∠ CAD = ∠ CMD
In this way, we can get EF / / ab

It is known that in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at point D, be bisection ∠ ABC, intersection ad at point m, an bisection ∠ DAC, intersection BC at point n Verification: the quadrilateral amne is rhombic

It is proved that: ∵ ad ⊥ BC,  BDA = 90 °, ∵∵ BAC = 90 °,  ABC + ∠ C = 90 °, ABC + ∠ bad = 90 °, ∵ bad = ∠ C, ? an bisection ? DAC, ? can = ∠ Dan, ? ban = ∠ bad + ∠ Dan, ? be ⊥ an

As shown in the figure, it is known that △ ABC, ∠ CAE is the external angle of △ ABC, in the following three items: ① AB = AC; ② ad bisection ∠ CAE; ③ ad ‖ BC. Select two items as the topic, and the other as the conclusion to form a true proposition and prove it

Proposition: if ① ② then ③. The proof is as follows:
∵AB=AC，
∴∠ABC=∠ACB．
∵ ad bisection ∵ CAE,
∴∠DAE=∠CAD．
And ∠ DAE + ∠ CAD = ∠ ABC + ∠ ACB,
∴2∠CAD=2∠C，
That is ∠ CAD = ∠ C,
∴AD∥BC．

In the triangle ABC, point D and point e are on the side BC, the angle CAE is equal to the angle B, e is the midpoint of CD, and the angle Bae is bisected by ad. (1) when the angle BAC = 90, it is proved that BD = AC; (2) if the angle BAC is not 90, whether BD = AC is tenable? Explain the reasons No similarity Just answer question 2~ (2) Angle ADC = angle B + angle bad Angle DAC = angle DAE + angle CAE = angle bad + angle B So, angle ADC = angle DAC, so CA = CD The bisector of AC through D intersects AC with F, It can be proved that △ ace and △ DCF are congruent, DF = AE You should have no problem working on it I still can't go down~

(1) Because angle BAC = 90, so angle B + angle c = 90, because angle B = angle CAE, so angle CAE + angle c = 90, so AE is perpendicular to CD, because e is the middle point of CD, AE vertically bisects CD, so ad = AC, angle CAE = angle DAE, so angle B = angle DAE because ad bisects angle BAE, angle bad = angle DAE, so angle B = angle bad, so triangle