As shown in the figure, B is a point on the line ad, △ ABC and △ BDE are equilateral triangles, which connect CE and extend the extension line crossing ad at point F, and the circumscribed circle ⊙ o of ⊙ ABC intersects CF at point P (1) It is proved that be is the tangent of ⊙ o; (2) If CP = 2, PF = 8, find the length of AC

As shown in the figure, B is a point on the line ad, △ ABC and △ BDE are equilateral triangles, which connect CE and extend the extension line crossing ad at point F, and the circumscribed circle ⊙ o of ⊙ ABC intersects CF at point P (1) It is proved that be is the tangent of ⊙ o; (2) If CP = 2, PF = 8, find the length of AC

(1) It is proved that: connected ob; ∵ ABC and △ BDE are equilateral triangles,  ABC = ∠ EBD = 60 °. CBE = 180 ° - 60 ° - 60 ° = 60 °. And ? OBC = 12  ABC = 30 °, and ? OBE = ∠ OBC + ∠ CBE = 90 °. OB ⊥ be. ? be is the tangent of ⊙ O. (2) connect AP, then ∵ APC = ∠ a

B is a point on the line ad, △ ABC and △ BDE are equilateral triangles, ⊙ q is the circumcircle of ⊙ ABC. CE and ⊙ o intersect at g. if de: BC = 1:2, find eg: CG

Let de = 1, BC = 2
(1) If points c and E are on both sides of AD, then point B coincides with G, eg: CG = EB: CB = de: BC = 1:2
(2) If point C and E are on the same side of AD, then ∠ CBE = 180 - ∠ ABC - ∠ DBE = 60
And be = de = BC / 2,
Therefore △ CBE is a right triangle, ∠ BEC = 90, ∠ BCE = 30, CE = bcsin ∠ CBE = √ 3
Therefore, ∠ GAC = ∠ BCE = 30, ∠ ACG = ∠ ACB + ∠ GAC = 90
Therefore, CG = actan30 = 1, Ge = ce-cg = √ 3-1
So eg: CG = √ 3-1

As shown in the figure, in △ ABC, ∠ C = 90 °, AC = BC, ad bisects ∠ cab, intersects BC at point D, de ⊥ AB at point E. if the circumference of △ BDE is 4cm, then the length of AB is______ cm．

∵ C = 90 °, de ⊥ AB, ad bisection ∠ cab,
∴CD=DE．
And ∵ ad = ad,
∴Rt△ACD≌Rt△AED（HL），
∴AC=AE，
∴BD+DE=BD+CD=BC．
And ∵ AC = BC,
∴AE=BC，
The circumference of △ BDE = BD + de + be = AE + be = 4cm,
∴AB=4cm．
Therefore, fill in 4

As shown in the figure, in △ ABC, the angle c = 90, AC = BC, the bisector ad of angle cab intersects BC at D, De is perpendicular to AB, if the circumference of triangle BDE is 4cm, find AB's I am more than benzene Please help; do it if you can Words in lesson 23.24 Thanks!very much! As shown in the figure, in △ ABC, the angle c = 90, AC = BC, the bisector ad of angle cab intersects BC at D, De is perpendicular to AB, if the perimeter of triangle BDE is 4cm, find the length of ab

AB = 4, because ad is the angular bisector of ∠ cab. According to the definition of angular bisector, CD = de proves that triangle ACD ≌ triangle AED can be obtained, AE = AC because AC = ab
So AE = BC, BC = CD + BD = de + BD, so AB = BD + de + be = 4

As shown in the figure, in △ ABC, ∠ C = 90 °, AC = BC, Da bisects ∠ cab and intersects BC with point D. can we determine a point E on AB so that the circumference of △ BDE is equal to the length of AB? If yes, please make e point and give proof; if not, please explain the reason

It is proved that a point E can be determined on AB such that the perimeter of △ BDE is equal to the length of ab. it is proved that the point D is de ⊥ AB at the point E, ∵ in △ ABC,

As shown in the figure, in △ ABC, ∠ C = 90 °, AC = BC, Da bisects ∠ cab and intersects BC with point D. can we determine a point E on AB so that the circumference of △ BDE is equal to the length of AB? If yes, please make e point and give proof; if not, please explain the reason

It is possible to determine a point E on AB such that the circumference of △ BDE is equal to that of ab
It is proved that the point D is de ⊥ AB at point E,
∵ in ᙽ ABC, ∵ C = 90 ° Da bisection ᙽ cab,
∴DC=DE，∠CDA=∠EDA，
∴AE=AC，
∵AC=BC，
∴∠B=45°，BC=AE，
The △ bed is an isosceles right triangle,
∴DE=BE，
The circumference of △ BDE is de + BD + be = DC + BD + be = BC + be = AE + be = ab

As shown in the figure, in △ ABC, ∠ C = 90 °, AC = BC, ad bisects ∠ cab, intersects BC at point D, de ⊥ AB at point E. if the circumference of △ BDE is 4cm, then the length of AB is______ cm．

∵ C = 90 °, de ⊥ AB, ad bisection ∠ cab,
∴CD=DE．
And ∵ ad = ad,
∴Rt△ACD≌Rt△AED（HL），
∴AC=AE，
∴BD+DE=BD+CD=BC．
And ∵ AC = BC,
∴AE=BC，
The circumference of △ BDE = BD + de + be = AE + be = 4cm,
∴AB=4cm．
Therefore, fill in 4

In the acute triangle ABC, the high AD and CE intersect at the point h, s △ BDE = 100, s △ BAC = 900 Find SINB

CE ⊥ AB, ad ⊥ BC, four points of a, e, D, C are in a circle, ⊥ bed = ﹤ BCA (the outer angle of a quadrangle inside a circle is equal to the inner diagonal) ﹤ a = < A, △ BDE ∷ BAC, s △ BDE / s ⊥ BAC = (BD / AB) ^ 2100 / 900 = (BD / AB) ^ 2, BD / AB = 1 / 3, CoSb = BD / AB = 1 / 3, SINB = √ [1 - (CoSb) ^ 2] = 2 √ 2 / 3

As shown in the figure: AD and CE are the height of triangle ABC, (1) Verification: △ BDE ∽ BAC; (2) If AC = 10, 5bd = 3ba, find the length of de

(1) It is proved that: ∵ ad, CE are the height of triangle ABC,
∴∠ADB=∠CEB=90°，
∵∠B=∠B，
∵△BDA∽△BEC，
∴BE
BD=BC
BA，
∵∠B=∠B，
∴△BDE∽△BAC
（2）∵△BDE∽△BAC，
∴DE
AC=BD
AB，
∵AC=10，5BD=3BA，
∴DE
10=3
5AB
AB，
De = 6

In the triangle ABC, DC = two thirds AC, CE = quarter BC, the area of triangle BDE is 24 square centimeter, what is the area of triangle ABC?

Do is set high to be through point D, and AQ to BC through point a
According to the meaning of the title:
SBDE=1/2*BE*DO=24
SABC=1/2*BC*AQ
Because: BC / be = 4 / 3
AQ/DO=3/2
so:SABC=1/2*(4/3BE)*(3/2DO)
=SBDE*4/3*3/2
=24*2
=48