Additional question: as shown in the figure, it is known that △ ABC is inscribed in ⊙ o, AB is the diameter, ∠ CAE = ∠ B Verification: AE and ⊙ o are tangent to point a

Additional question: as shown in the figure, it is known that △ ABC is inscribed in ⊙ o, AB is the diameter, ∠ CAE = ∠ B Verification: AE and ⊙ o are tangent to point a

It is proved that: ∵ AB is the diameter,
∴∠ACB=90°，
∴∠BAC+∠B=90°，
And ∵ CAE = ∵ B,
∴∠BAC+∠CAE=90°，
In other words, BAE = 90 °,
So AE and ⊙ o are tangent to point a

As shown in the figure, triangle ABC is inscribed in circle O, AB is the diameter of circle O, CD bisects ∠ ABC, intersects circle O at point D, intersects AB at point F, chord AB is perpendicular to point h, connecting CE and oh,

⊙ o in F ∵ AB is the diameter of the circle O

It is known that in △ ABC, ∠ BAC = 90 °, ab = AC, AE is a straight line passing through point a, and points B and C are on the opposite side of AE, BD ⊥ AE is at point D, CE ⊥ AE is at point E Explanation: BD = de + CE

BD⊥AE,CE⊥AE
Then BD / / CE, ∠ DBC = ∠ BCE
AB = AC, then ∠ ACB = ∠ abd + ∠ DBC = 45 degrees
RT triangle ace
∠EAC=90-∠ACB-∠BCE=45-∠BCE=45-∠DBC=∠ABD
AB = AC
So the congruence of cart and the triangle of arc
Ad = CE, BD = AE
Because AE = AD + de
So BD = AE = AD + de = CE + De

Additional question: as shown in the figure, it is known that △ ABC is inscribed in ⊙ o, AB is the diameter, ∠ CAE = ∠ B Verification: AE and ⊙ o are tangent to point a

It is proved that: ∵ AB is the diameter,
∴∠ACB=90°，
∴∠BAC+∠B=90°，
And ∵ CAE = ∵ B,
∴∠BAC+∠CAE=90°，
In other words, BAE = 90 °,
So AE and ⊙ o are tangent to point a

As shown in the figure, △ ABC is connected to ⊙ o, AB is the diameter of ⊙ o, CD bisection ⊙ ACB intersects ⊙ o at point D, AB intersects AB at point F, chord AE ⊥ CD at point h, connecting CE and oh (1) Verification: △ ace ∽ CFB; (2) If AC = 6, BC = 4, find the length of oh

(1) Prove that: ∵ AB is the diameter of ⊙ o,
∴∠ACB=90°；
∵ CD bisection ∵ ACB,
∴∠ACD=∠FCB=45°；
∵AE⊥CD，
∴∠CAE=45°=∠FCB；
In △ ace and △ BCF, ∠ CAE = ∠ FCB, ∠ e = ∠ B,
∴△ACE∽△CFB；
(2) Extend AE and CB at point M;
∵∠FCB=45°，∠CHM=90°，
∴∠M=45°=∠CAE；
∴HA=HC=HM，CM=CA=6；
∵CB=4，
∴BM=6-4=2；
∵OA=OB，HA=HM，
/ / Oh is the median line of △ ABM,
∴OH=1
2BM=1．

In the isosceles triangle ABC, ab = AC, ∠ BAC = 120 °, D is the point on BC, and BD = 1, DC = 2, find ad

Pass a as AE ⊥ BC to E,
By ∠ BAC = 120 °, B = ∠ C = 30 °,
From CD = 2, BD = 1,  CE = 3 / 2
∴DE=3/2-1=1/2,
From ∠ DAE = 120 ° to 2-30 ° = 30 °,
∴AD=BD=1.

Inquiry 1: as shown in the figure, e is any point on AB side and △ CDE is an equilateral triangle, connecting ad, conjecturing the positional relationship between AD and BC, and explain the reasons Inquiry 2: as shown in the figure, if △ ABC is any isosceles triangle, ab = AC, e is any point of AB, △ CDE is isosceles triangle, de = DC, and ∠ BAC = ∠ EDC, connect ad, guess the position relationship between AD and BC, and explain the reasons

(1) The position relationship between AD and BC is ad ∵ BC; ∵ ABC and △ Dec are equilateral triangles, ∵ ABC ∵ Dec, ∵ ACB = DCE = 60 °. ACBC = DCEC, ∠ DCA = ∠ ECB. ∵ ACD ﹤ BCE. ? DAC = ∵ ACB. ∵ ad ∽ BC. (2) the position relationship between AD and BC is

In the circle O, two chords AB and CD intersect at point P, and ab = CD. It is proved that PA = PC, Pb = PD?

Proof: connect AC and BD
The arc corresponding to ∵ cab and  CDB is arc BC
∴∠CAB＝∠CDB
∵∠APC＝∠DPB
The △ APC is similar to △ DPB
∴PA/PC＝PD/PB
Qi PA.PB=PC .PD

As shown in the figure, the extension lines of chord AB and CD of circle O intersect with P, and PA = PC. It is proved that Pb = PD

There is a far cut string theorem PA * Pb = PD * PC
Because PA = PC, Pb = PD

Known: as shown in the figure, AB is the diameter of ⊙ o, C, D are two points on ⊙ o, and C is For the midpoint of AD, if ∠ bad = 20 °, calculate the degree of ∠ ACO

∵ AB is the diameter of ⊙ o, and C is
The midpoint of AD,
∴OC⊥AD，
∵∠BAD=20°，
∴∠AOC=90°-∠BAD=70°，
∵OA=OC，
∴∠ACO=∠CAO=180°−∠AOC
2=180°−70°
2=55°．