As shown in the figure, in △ ABC, D is a point on AC, e is a point on CB extension line, and AC BC＝EF DF， Confirmation: ad = EB

As shown in the figure, in △ ABC, D is a point on AC, e is a point on CB extension line, and AC BC＝EF DF， Confirmation: ad = EB

It is proved that DH ∥ BC is made through point D and ab is crossed with H, as shown in Fig,
∵DH∥BC，
∴△AHD∽△ABC，
∴AD
AC=DH
BC is ad
DH=AC
BC，
∵DH∥BE，
∴△BEF∽△HDF，
∴BE
HD=EF
DF，
And AC
BC＝EF
DF，
∴BE
HD=AD
DH，
∴AD=EB．

As shown in the figure, △ ABC, ab = AC, e is a point on AB, f is a point on AC extension line, and be = CF, if EF and BC intersect at D, prove: de = DF

It is proved that FH is an extension of ab,
∵FH∥AB，
∴∠FHC=∠B．
And ∵ AB = AC,
∴∠B=∠ACB．
And ∵ ACB = ∵ FCH,
∴∠FHE=∠FCH．
∴CF=HF．
And ∵ be = CF,
∴HF=BE．
And ∵ FH ∥ ab,
∴∠BED=∠HFD，
In △ DBE and △ fhe,
∠B＝∠FHC
BE＝HF
∠BED＝∠HFD ，
∴△DBE≌△FHE（ASA）．
∴DE=DF．

As shown in the figure, AB is the diameter of ⊙ o, EF is the chord, CE ⊥ EF intersects AB with C, DF ⊥ EF intersects AB with D. verification: AC = BD

O for og ⊥ EF for EF to g
∵ EF is the chord of ⊙ o, and og ⊥ EF, ᙽ eg = FG
∵ CE ⊥ EF, DF ⊥ EF, og ⊥ EF, ᙽ og ⊥ CE ∥ DF,  cdfe is trapezoidal,
Combined with the obtained eg = FG, it is concluded that og is the median line of trapezoidal cdfe and ﹥ OC = OD
Obviously, there are: OA = ob, OA OC = ob OD, AC = BD

It is known that AB is the diameter of circle O, CD is the chord, be ⊥ CD is in E, AF ⊥ CD is in F, connecting OE and of （1）OE=OF； （2）CE=DF．

(1) It is proved that: connect OC, OD, og, make oh ⊥ BG in H, cross CD in M, ∵ AB is the diameter of circle O, be ⊥ CD in E, AF ⊥ CD in F, ∵ BGF = 90 °, Quad BGFE is a rectangle, ᚉ BG = EF, BG ∥ EF, ∵ oh ⊥ BH = GH, EF ⊥ Oh, ᙽ bhme and ghmf are also rectangles

Let CD, o be the extension of a circle 1. Try to judge the shape of the triangle OEF and explain the reasons: 2. Prove: arc AC = arc BD!

isosceles triangle
ac=bd
Prove with congruence
It's easy

As shown in the figure, in △ ABC, ab = AC, the circle O with ab as its diameter intersects BC at D and AC at point E. DF ⊥ AC is made through D, the perpendicular foot is f (1), and DF is the tangent line of circle o

Because the circle O with diameter AB intersects BC and D,
So ad ⊥ BC,
Because AB = AC,
So BD = CD,
Ao = Bo
So OD ‖ AC
Because DF ⊥ AC
So OD ⊥ DF
So DF is tangent to circle o

AB is the diameter of circle O, EF is the chord, CE ⊥ EF, DF ⊥ EF, e and F are perpendicular feet prove: It is om ⊥ EF when passing through the center of circle O, and M is perpendicular foot Then according to the "vertical diameter theorem", me = MF Because CE ⊥ EF, DF ⊥ EF So CE / / OM / / DF So OC / OD = me / MF = 1 OC = so Because OA = ob So AC = BD Why can OC / OD = me / MF = 1 be introduced

Because OC = od = R, that is the radius of the circle
Because EF is the chord of a circle, and O is the center of the circle. If a chord is perpendicular to the center of the circle, the vertical line bisects the chord!
(if OE and of are connected, then OE = of, which is radius, so △ OEF is an isosceles triangle, so the bottom edge is bisected by height.)

In the circle O, the diameter AB intersects with the chord CD, passing through three points a, O and B respectively to make the vertical lines of CD, and the vertical feet are e, h and f respectively. Verification: CE = DF

Let the intersection point of AB and CD be g. according to the similarity relation, BG / FG = og / GH = OA / EH,
So (BG + OG) / (FG + GH) = OA / eh = > FG = GH, h is the midpoint of BC, so CE = DF

As shown in the figure, AB is the diameter of circle O, ad is the chord, e is a point outside circle O, EF is perpendicular to F, intersecting ad at point C, and CE = ed. it is proved that De is the tangent line of circle o

prove:
Connect OD
∵OD=OA
∴∠ODA=∠A
∵EC=ED
∴∠EDC=∠ECD=∠ACF
∵EF⊥AB
∴∠A+∠ACF=90°
∴∠ADO+∠CDE=90°
That is OD ⊥ De
/ / De is the tangent of circle o

As shown in the figure, AB is the diameter of ⊙ o, AC is the chord, the straight line CE and ⊙ o are tangent to point C, ad ⊥ CE, and the perpendicular foot is d

Proof: connect BC,
∵ AB is the diameter of ⊙ o,
∴∠ACB=90°，
∴∠B+∠CAB=90°；
∵AD⊥CE，
∴∠ADC=90°，
∴∠ACD+∠DAC=90°；
∵ AC is the chord, and CE and ⊙ o are tangent to point C,
∴∠ACD=∠B，
Ψ DAC = ∠ cab, i.e. AC bisection ∠ bad