As shown in the figure, AB is the diameter of ⊙ o, C is ⊙ o, and the tangent line passing through point C is perpendicular to each other, and the perpendicular foot is D, so as to verify the AC bisection angle DAB As shown in the figure, AB is the diameter of ⊙ o, C is the point on ⊙ o, CD cuts ⊙ o at point C, [label: ab CD, diameter, CD]

As shown in the figure, AB is the diameter of ⊙ o, C is ⊙ o, and the tangent line passing through point C is perpendicular to each other, and the perpendicular foot is D, so as to verify the AC bisection angle DAB As shown in the figure, AB is the diameter of ⊙ o, C is the point on ⊙ o, CD cuts ⊙ o at point C, [label: ab CD, diameter, CD]

∵OC⊥CD,AD⊥CD
∴OC‖AD
∴∠OCA=∠CAD
And ∵ Ao = Co
∴∠OCA=∠CAO
∴∠CAD=∠CAO
ν AC bisection angle DAB

As shown in the figure, AB is the diameter of O and C is the point on the center O. the tangent lines between AD and point C are perpendicular to each other, and the perpendicular foot is D. the AC bisection angle DAB is verified

OC ⊥ CD,
Ad ⊥ CD,
∴AD‖OC,
∴∠DAC=∠ACO,
∵OA=OC,
∴∠CAO=∠ACO=∠DAC,
That is, AC bisection ∠ DAB

As shown in the figure, AB is the diameter of ⊙ o, C is the point on ⊙ o, the tangent lines of AD and point C are perpendicular to each other, and the perpendicular foot is d (1) Verification: AC bisection ∠ DAB; (2) If ad = 4, AC = 5, find ab

(1) It is proved that: connecting OC, ∵ C is the point above ⊙ o, DC is tangent, ∵ OC ⊥ CD. And ∵ ad ⊥ DC, ∵ ad ? OC, ∵ DAC = ∵ ACO. And ? Ao = OC, ? Cao = ∠ ACO, ? DAC = ∠ Cao

As shown in the figure, it is known that AB is the diameter of ⊙ o, and point C is The middle point of AE is the chord CD ⊥ AB through C, and AE is intersected with F. it is proved that AF = CF

Proof: connect AC,
∵ chord CD ⊥ AB, AB is the diameter of ⊙ o,
Qi
AC=
AD，
∵ point C is
The midpoint of AE,
Qi
AC=
CE，
Qi
AD=
CE，
∴∠ACD=∠CAE，
∴AF=CF．

As shown in the figure, AB is the diameter of circle O, AE is the chord, EF is the tangent, e is the tangent point, AF is vertical EF, and the perpendicular foot is f

Proof: link OE,
Because EF is the tangent of circle o,
So OE is perpendicular to ef,
Because AF is perpendicular to EF,
So OE / / AF,
So angle AEO = angle FAE,
Because OA = OE,
So angle AEO = angle OAE,
So the angle FAE = the angle OAE,
So AE bisection angle fab

As shown in the figure, AB is the diameter of circle O, chord CD ⊥ AB is at point m, chord AF intersects CD at point e. it is proved that the square of AC = AE times AF

According to the projective theorem, we know that the square of AC = am times AB, and then according to the triangle AEM similar to the triangle ABF (this should prove similar, there is a same angle and two right angles), find AE / AB = am / AF, deduce, AE * AF = AB * am, so, equivalent substitution, get the proof!

As shown in the figure, AB is the chord of circle O. the circle with a as its center intersects with circle O at C, D, AB with E, CD with AB with F. it is proved that AE? 2 = AF · ab The level of the picture is not enough. It's not easy to pass on. Draw it by yourself

If AB is the diameter of circle O, BD is the chord of ⊙ o, extend BD to point C, BD = CD; connect AC, pass point D as de ⊥ AC, perpendicular foot is e (1) verification: ab = AC (2) verification: De is tangent of ⊙ o (3) if radius of ⊙ o is 5, ﹤ BAC = 60 °, find the length proof of de: because BD = CD, and OA = ob

The diameter ab of the circle O is perpendicular to the chord CD, f is a point on the extension line of CD, connecting with the intersection point of circle O and e. it is proved that AC square = AE * AF

0

As shown in the figure, AB is the diameter of semicircle, CD ⊥ AB is at D, chord AF intersects CD at e, and intersects semicircle at point F. if CE = AE, it is proved that C is the midpoint of arc AF

Proof: even BC, BF
Because AB is the diameter, CD ⊥ AB is in D
So CD is the height on the hypotenuse of the right triangle ABC
So ∠ ACD = ∠ ABC
Because CE = AE
So ∠ ACD = ∠ fac
So ∠ ABC = ∠ fac,
Because ∠ fac = ∠ FBC,
So ∠ FBC = ∠ ABC
So C is the midpoint of the arc AF

In the triangle ABC, the angle BAC = 90, ab = AC, point E is on the extension line of side BC, Da is vertical AE, ad = AE, point F is the midpoint of De, and CF = DF is proved RT

Proof: connect CD
In △ ADC and △ AEB, because ∠ CAE = 90 ° + ∠ CAE = ∠ BAE, ad = AE, AC = ab
So: these two triangles are congruent
So: ∠ ADC = ∠ AEB, i.e. ∠ ADC = ∠ AEC
Therefore: A, D, e, C four points in the garden
So: ∠ DCE = ∠ DAE = 90 °
And F is the midpoint of de,
So: DF = CF