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As shown in the figure, in △ ABC, D is the midpoint of BC, de ⊥ BC, the bisector of ⊥ BAC is AE at point E, EF ⊥ AB is at point F, eg ⊥ AC intersects AC extension at point g. verification: BF = CG
Proof: connect be, EC,
∵ED⊥BC，
D is the midpoint of BC,
∴BE=EC，
∵EF⊥AB EG⊥AG，
AE ∠,
∴FE=EG，
In RT △ BFE and RT △ CGE,
BE＝CE
EF＝EG ，
∴Rt△BFE≌Rt△CGE （HL），
∴BF=CG．
As shown in the figure, in △ ABC, D is the midpoint of BC, de ⊥ BC, the bisector of ⊥ BAC is AE at point E, EF ⊥ AB is at point F, eg ⊥ AC intersects AC extension at point g. verification: BF = CG
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Draw a circle with P as the center and Pb as the radius, intersect AP with D and connect BD
Then: △ PBE is a regular triangle
That is: PD = Pb
∵∠ADB=180-60=120º,∠CPB=60+60=120º
∴∠ADB=∠CPB ①
∵ bad = ∵ BCP (corresponding to the same arc) ②
From ① and ② we can get: ∠ abd = ∠ CBP ③
∵ △ ABC is an equilateral triangle
∴AB=CB ④
From the synthesis of (2), (3) and (4), it is concluded that: 1
∴△ABD≌△CBP（ASA）
∴AD=CP
∴PA=PD+AD=PB+PC
As shown in the figure, P is an equilateral △ ABC circumscribed circle At any point on BC, it is proved that PA = Pb + PC
It is proved that: intercept PD = PC, ∵ AB = AC = BC, ? APB = ∵ APC = 60 degrees, ? PCD is an equilateral triangle, ? PCD = ∠ ACB = 60 °, CP = CD, ? PCD - ? DCB = ? ACB - ? DCB, ? ACD = AC = BC in △ ACD and △ BCP, AC = BC ﹤ ACD ﹤ CP = CD ﹣ and
Δ ABC is the inscribed equilateral triangle of ⊙ o, P is This paper explores the quantitative relationship between PA and Pb + PC, and explains the reasons
PA=PB+PC．
The reasons are as follows: as shown in the figure, from the circular angle theorem, ∠ BAE = ∠ BCP, ∠ APB = ∠ ACB = 60 °,
If PE = BP is truncated on PA, then △ PBE is an equilateral triangle,
So, Pb = be,
∵∠AEB=180°-60°=120°，
∠CPB=120°，
∴∠AEB=∠CPB=120°，
∵ △ ABC is an equilateral triangle,
∴AB=BC，
In △ Abe and △ CBP,
∠BAE＝∠BCP
∠AEB＝∠CPB
AB＝AC ，
∴△ABE≌△CBP（AAS），
∴AE=PC，
∵PA=AE+PE，
∴PA=PB+PC．
It is known that the triangle ABC is an equilateral triangle and P is a point on the circumcircle of the triangle ABC. When p is on the arc BC, it is proved that PA = Pb + PC It should be explained in detail
If BD is parallel to PC, then ∠ DBC = ∠ BCP, because ∠ BPC = 120 degrees, so ∠ PBC + ∠ PCB = 60 degrees, so ∠ PBC + ∠ DBC = 60 degrees, because ∠ ABC = 60 degrees, so ∠ abd = ∠ CBP, because ∠ BCP = ∠ BAP, and ab = BC, so △ ADB ≌ △ BPC,
Because BD = BP (congruence) and ∠ DBP = 60 degrees, so BD = DP, and because PC = ad, so AP = AD + DP = BP + PC! This is the most convenient solution at present!
As shown in the figure, P is a point inside the equilateral △ ABC, and PA = 6, PC = 8, Pb = 10, D is the outer point of △ ABC, and △ ADC ≌ △ APB, calculate the degree of ∠ APC
As shown in the figure,
Connect DP,
∵ △ ABC is an equilateral triangle,
∴∠BAC=60°，
∵△ADC≌△APB，
∴∠DAC=∠PAB，DA=PA，DC=PB，
∵∠PAC+∠BAP=60°，
∴∠PAC+∠CAD=60°，
The △ DAP is an equilateral triangle,
∴DP=6，∠DPA=60°；
In △ PDC
PC=8，DP=6，DC=10，
∵82+62=102，
∴∠DPC=90°，
∴∠APC=∠DPA+∠DPC=60°+90°=150°．
As shown in the figure, we know that the equilateral triangle ABC, P is the point outside the triangle ABC, and the angle APC is equal to 60 degrees. It is proved that pa-pc = Pb
On AP, take PD = PC, connect CD, < DPC = 60 degrees, PD = PC, triangle PCD is equilateral triangle, CD = PC, AC = BC, < ACD = < ACB - < DCB, < BCP = < DCP - < DCB, < ACB = < DCP = 60 degrees, < ACD = < BCP, △ ACD ≌ △ BCP, BP = ad,
∴AP=AD+PD=PC+BP,
That is: pa-pc = BP
In the triangle ABC, if the vertical bisector of sides AB and AC intersects at point P, then the relationship among PA, Pb and PC is? Please write down the process of solving the problem
PA=PB=PC
The distance between the point on the vertical bisector of a line segment to both ends of the line segment is equal
PA=PB,PB=PC
For example, if the vertex of a, B, B, a, B, a, B, a, B, a, B, a, B, B, a, B, B, a, B, B, a, B, a, B, B, a, B, B, a, B, B, a, B, a, B, B, a, B, a, B, B, B, B, B, B
Even PA, as shown in the figure,
∵ the vertical bisector of AB and AC intersects at point P,
∴AP=BP，AP=CP，
∴∠1=∠3，∠2=∠4，
And ∠ a = 70 °,
∴∠3+∠4=∠1+∠2=70°，
And ∠ 3 + ∠ 5 + ∠ 4 + ∠ 6 = 180 ° - 70 ° = 110 °,
∴∠5+∠6=110°-70°=40°，
∴∠BPC=180°-∠5-∠6=140°．
RELATED INFORMATIONS
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