AB is the chord of circle O, AE, arc = BF, arc radius OE, of intersect a, B, C, D respectively, which indicates that OCD is isosceles triangle

AB is the chord of circle O, AE, arc = BF, arc radius OE, of intersect a, B, C, D respectively, which indicates that OCD is isosceles triangle

Proof: connect AO and Bo
Because: AE arc = BF arc
So: angle AOC = angle BOD
Because: Ao = Bo
So: angle OAC = angle OBD
So: Triangle OAC is equal to triangle OBD
So: OC = OD
So: Triangle OCD is isosceles triangle

It is known that the radius OC and OD of circle O intersect with points E and f respectively at chord AB, and arc AC = arc BD. it is proved that AE = BF

Connection: OA, ob
Let the radius of the circle be r
Then: OA = ob = OC = od = R
So: angle ABO = angle Bao
Because: arc AC = arc BD
So: angle AOC = angle BOD
Angle AFO = angle ABO + angle BOD
Angle BeO = angle Bao + angle AOC
Angle AFO = angle BeO
OE = of, Ao = Bo, angle AOC = angle BOD
Triangle AOE and triangle BOF are congruent triangles
AE=BF

Let B be a graph with the length of a B B = 12, if B B B = B B C = B B C Come on

Connecting AF
AB=AC,∠BAC=120
∠B=∠BAF=30°
∠FAC=90
∠C=30°
Let BF = x = AF, CF = 2x
x+2x=12
X=4

As shown in the figure △ ABC, ab = AC, ∠ BAC = 120 ° EF is the vertical bisection of AB, EF intersects BC at f and ab intersects AB at point e. it is proved that BF = half FC

∵AB=AC
∴∠ABC=∠ACB=30°
∵ EF vertical bisection ab
∴BF=AF
∠BAF=∠ACF=30°
∴∠FAC=90°
In RT △ AFC, ∠ C = 30 °
∴AF=（1/2）CF
BF = AF
∴BF=（1/2）CF

As shown in the figure, in △ ABC, point D is on BC and CD = Ca, CF bisection ∠ ACB, AE = EB, verification: EF = 1 2BD．

It is proved that ∵ CD = Ca, CF bisection ∠ ACB,
/ / F is the mid point of AD,
∵AE=EB，
Ψ e is the midpoint of ab,
∴EF=1
2BD．

In the triangle ABC, CD / DA = AE / EB = 1 / 2, vector BC = vector a, vector CA = vector B, prove that vector de = 1 / 3 (vector b-vector a)

DE=AE-AD=(1/3)AB-(2/3)AC=(1/3)(-a-b)-(2/3)(-b)=(1/3)(b-a)

The triangle ABC angle c = 90, D is the midpoint of AB, the vertical DF E F is the point on Ca CB respectively. It is proved that AE square and BF square are equal to the square of EF

It is proved that if FD is extended to G so that FD = DG and Ag is connected, then: △ ADG ≌ △ BDF, so: BF = AG, FD = DG, ∠ DBF = ∠ DAG, so: Ag ≌ BC, de vertically bisect FG, so: ∠ gae = 90 °, EF = eg, so: Ag ≌△ BDF in RT △ AEG, i.e., FB ≌ + AE ≌ = ef

In △ ABC, CA = CB, ∠ C = 90 ° D is any point on AB, AE ⊥ CD, BF ⊥ CD, and the result is: EF = ae-bf │

1)AD

It is known that in △ ABC, CA = CB, ∠ C = 90 ° D is any point on AB, AE ⊥ CD, perpendicular foot is e, BF ⊥ CD, and vertical foot is F. it is proved that EF = | ae-bf |

Proof: AE ⊥ CD,
∴∠AEC=90°，
﹤ ACE + ∠ CAE = 90 °, (two acute angles of a right triangle complement each other)
∵∠ACE+∠BCF=90°，
Ψ CAE = ∠ BCF, (the remainder of equal angle is equal)
∵AE⊥CD，BF⊥CD，
∴∠AEC=∠BFC=90°，
And △ in CBF,
∠AEC＝∠BFC
∠CAE＝∠BCF
AC＝BC
∴△ACE≌△CBF（AAS），
∴AE=CF，CE=BF，
∴EF=CE-CF=BF-AE，
When AE > BF, as shown in Fig,
EF = ae-bf can be obtained by the same method,
That is, EF = | ae-bf |

It is known that in the triangle ABC, CA = CB, angle c = 90 degrees, D is the point above AB, AE vertical CD, BF vertical CD, verification: EF = ae-bf

∵∠C=90°
∴∠ACD+∠BCD=90°
∵AE⊥CD
∴∠EAC+∠ACD=90°
∴∠BCD=∠EAC
∵CA=CB
∴RT△CEA≌RT△BCF（AAS）
∴CE=BF AE=CF
∵CF-CE=EF
ν ae-bf = ef (substituted)
Do you understand?