As shown in the figure, in the equilateral triangle ABC, the bisectors of ∠ B and ∠ C intersect at O, and the vertical bisectors of OB and OC intersect BC at e and F. try to explore the size relationship of be, EF and FC, and explain the reasons

As shown in the figure, in the equilateral triangle ABC, the bisectors of ∠ B and ∠ C intersect at O, and the vertical bisectors of OB and OC intersect BC at e and F. try to explore the size relationship of be, EF and FC, and explain the reasons

Conclusion: be = EF = FC (1 point) the reason is: be = EF = FC (1 point) the reason is: ∵ ABC is equilateral triangle, ∵ ABC = ∠ ACB = 60 ° (2 points), ∵ OC, OB bisection ∵ ACB, ∵ ABC, ∵ OBE = ∠ OCF = 30 ° (3 points), ∵ eg, HF vertical bisection ob, OC, ∵ OE = be, of = FC (5 points), ∵ BOE = ∠ OBE = 30 °, ∵ COF = ∠ OCF = 30 °, ∵ OEF = ∠ ofe = 60 °, ∵ triangle OEF is equilateral triangle (8 points), Of = OE = EF, be = EF = FC (10 points)
As shown in the figure, in the equilateral triangle ABC, the bisectors of ∠ B and ∠ C intersect at O, OB and OC, and the vertical bisectors BC intersect at e and F. it is proved that be = EF = FC
∵△ ABC is an equilateral triangle, with ∠ ABC ＝ ACB = 60 °. OBC ＝ ABC / 2, ∠ OCB ＝ ACB / 2, ∠ OBC ＝ OCB = 30 °. ∵ e is on the vertical bisector of ob, ∵ be = EO, ∵ OBC ＝ EOB, ∵ OEF = 2 ∠ OBC
What about the picture?
It is proved that because the triangle ABC is an equilateral triangle, the angle ABC and the angle ACB are both 60 ° and because ob OC is an angular bisector, the angle OBC and the angle OCB are both 30 ° and because OB and OC have a vertical bisector BC at e and F
It is proved that ∵ △ ABC is an equilateral triangle,
∴∠ABC=∠ACB=60°
∵ OC, ob, ABC, ∠ ACB,
∴∠OBE=∠OCF=30°
∵ eg, HF, ob, OC,
∴OE=BE，OF=FC
∴∠BOE=∠OBE=30，∠COF=∠OCF=30，
∴∠OEF=∠OFE=60°，
The triangle OEF is an equilateral triangle
∴OF=OE=EF，
What is the integral of the integral sign ((cos2x) / (COS ^ 2x) / (sin ^ 2x))
∫ cos2x / (sin²x * cos²x) dx
= ∫ cos2x / (1/2 * sin2x)² dx
= 4∫ cos2x / (sin²2x) dx
= 4∫ csc2x * cot2x dx
= -2∫ csc2x * cot2x d(2x)
= -2csc2x + C
= -2/(sin2x) + C
= -secx*cscx + C
As shown in the figure, in the triangle ABC, ab = AC = 6, P is any point on BC, please use the knowledge to find the value of PC times Pb plus PA square
Using Pythagorean theorem
Through the point, make Ao ⊥ BC, intersect BC at the point o. ∵ AB = AC = 6 ⊥ Bo = Co
The results of Pythagorean theorem are as follows
PC×PB+PA²
=（CO+OP)(CO-OP)+PA²
=CO²-OP²+PA²
=PA²-OP²+CO²
=AO²+CO²
=AC²
=36
On the application of sin ^ 2x + cos ^ 2x
Formula: cos ^ 6x + sin ^ 6x = (COS ^ 2x + sin ^ 2x) (COS ^ 4x cos ^ 2xsin ^ x + sin ^ 4x) = (COS ^ 2x + sin ^ 2x) ^ 2-3sin ^ 2xcos ^ 2x the specific idea PS: cos ^ 2x = the quadratic power of cosx
Cos ^ 6x + sin ^ 6x = (sin ^ 2x + cos ^ 2x) ^ 3-3sin ^ 2xcos ^ 2x (sin ^ 2x + cos ^ 2x) = 1-3sin ^ 2xcos ^ 2x makes full use of sin ^ 2x + cos ^ 2x = 1
In △ ABC, ab = AC = 6, P is any point on BC, try to find the value of (PC × Pb) + PA ^ 2
Let PC > Pb pass point a and make ad perpendicular to BC and D. from △ ABC is an isosceles triangle, BD = CDPA ^ 2 = ad ^ 2 + PD ^ 2 = AB ^ 2-bd ^ 2 + (pc-cd) ^ 2 = AB ^ 2-bd ^ 2 + PC ^ 2-2pc × CD + CD ^ 2 = AB ^ 2 + PC ^ 2-PC × BC = AB ^ 2 + PC (bc-pb) - PC × BC = AB ^ 2-PC × Pb, that is pa ^ 2 + PC × Pb = AB ^ 2 = 36
Set PC > Pb
Through point a, ad is perpendicular to BC and D, which can be seen from the isosceles triangle of △ ABC;
BD=CD
PA^2=AD^2+PD^2=AB^2-BD^2+（PC-CD）^2=AB^2-BD^2+PC^2-2PC×CD+CD^2
=AB^2+PC^2-PC×BC=AB^2+PC(BC-PB)-PC×BC=AB^2-PC×PB
That is: PA ^ 2 + PC × Pb = AB ^ 2 = 36