As shown in the figure, the known point O is a point on the inclined edge AC of RT △ ABC, and ⊙ o with point o as the center of the circle and the length of OA as the radius is tangent to point E, intersect with AC at point D, and connect AE (1) Verification: AE bisection ∠ cab; (2) This paper explores the quantitative relationship between ∠ 1 and ∠ C, and finds the value of Tanc when AE = EC

(1) Proof: connect OE,
⊙ O and BC are tangent to point E,
∴OE⊥BC，
∵AB⊥BC，
∴AB∥OE，
∴∠2=∠AEO，
∵OA=OE，
∴∠1=∠AEO，
Ψ 1 = ∠ 2, that is, AE bisection ∠ cab;
（2）∠C=90°-2∠1，tanC=
Three
3．
∵ EOC is the external angle of ∵ AOE,
∴∠1+∠AEO=∠EOC，
∵∠1=∠AEO，∠OEC=90°，
∴∠C=90°-2∠1，
When AE = CE, ∠ 1 = ∠ C,
∵2∠1+∠C=90°
∴3∠C=90°，∠C=30°
∴tanC=tan30°=
Three
3．

As shown in the figure, the known point O is a point on the hypotenuse AC of the RT triangle ABC. The circle with o as the center and OA length as the radius is tangent to point E, intersecting with AC at point D, and connecting AE (1) Verification: AE bisection angle cab; (2) The quantitative relationship between angle 1 and angle C in the graph is explored, and the value of Tanc when AE = CE is obtained

(1) In the triangle AOE, because OA = OE, angle OAE = angle OEA. Because BC is tangent to circle O, so OE is perpendicular to BC, then angle BAE = angle OEA, so angle BAE = angle OAE, then AE bisects angle cab
(2) No, where's corner 1

Given that the extension line of AB ed in circle O intersects C point angle c = 42 degrees arc BD = 30 degrees, calculate the degree of arc AE

This problem needs to use triangle interior angle sum theorem, quadrilateral angle sum theorem, radian = center angle angle degree = 2 times of circumference angle these knowledge

As shown in the figure, AB is the diameter of circle O and OD is parallel to ac. what is the relationship between arc CD and the size of arc BD? Why?

Connected to CO, DB
∠AOD+∠OAC+∠OCA=180
∵AO=CO
∴∠CAO=∠ACO
∵AC‖OD
∴∠ACO=∠COD
∠AOC+∠COD+∠DOB=180°
∴∠AOC+2∠DOB=∠AOB=180°
r*∠AOC+r*2∠DOB=r*∠AOB
Arc AC + 2 * arc DB = arc AB = semicircle

As shown in the figure, AB is the diameter of circle O. if od / / AC, what is the relationship between arc CD and arc BD, and why? [2.] exchange the condition in [1] with the conclusion Can it still hold? Explain why

(1) Arc CD = arc BD proof: connect and extend co intersection circle O with e because of OD / / AC, so ∠ DOA = ∠ OAC, ∠ C = ∠ DOE because OA = OC, so ∠ OAC = ∠ C, so ∠ DOA = ∠ BOE, so ∠ AOC + ∠ DOA = ∠ BOE + Doe, so ∠ cod = ∠ BOD, so arc CD = arc BD (2) still holds

AB is the diameter of circle O, OD is parallel to AC, what is the relationship between arc CD and arc BD

As shown in the figure:
Connect OC
∠OAC=∠OCA
∵OD‖BD
∴∠OCA=∠COD
∠OAC=∠BOD
∴∠COD=∠BOD
/ / arc CD = arc BD (in the same circle, the arcs opposite to the same center angle are equal)

In the circle O, if the chord AB and CD intersect at e, the angle AOC = 30, the angle BOD = 60, then the angle AEC degree

When ad is connected, then: ∠ AEC = ∠ ade ＋dae = (1 / 2) AOC ＋ (1 / 2) ∠ BOD = (1 / 2) ∠ AOC ∠ BOD) = (1 / 2) (30 ° + 60 °) = 45 °

AB is the diameter of the circle O and CD is the chord. The extension line of AB and CD intersects at the point e. it is known that Bo = De, ∠ AOC = 108 ° and the degree of ∠ e is obtained

BO=DE=DO,∠E=∠DOB
∠ODC=2∠E
OC=OD, ∠ODC=∠OCD=2∠E
∠DOC=180-4∠E
∠DOC+∠DOB=180-3∠E=180-108°
∠E=108°/3=36°

As shown in the figure, AB, AC are the two chords of ⊙ o, and ab = AC, D is On BC, P is If ∠ BDC = 150 °, find the degree of ∠ APC

In the inscribed quadrilateral ABCD, ∠ BAC = 180 ° - ∠ BDC = 180 ° - 150 ° = 30 °,
Then the degree of arc BC is 60 degrees,
And ∵ AB = AC,
The arc AB = arc AC = 150 °,
The arc ABC is 210 degrees,
∴∠APC=1
2×210°=105°．

As shown in the figure, AB, AC are the two chords of circle O, and ab = AC, D is a point on arc BC, P is a point on arc AC, if ∠ BDC = 150 °, calculate the degree of ∠ APC I've been preparing for it recently. I'm not sure,

∵ four points of a, B, D and C in a quadrilateral abdc are in a circle  BAC + ∠ BDC = 180  BAC = 180 - ∠ BDC = 180-150 = 30 ? AB = AC  ABC ＝ ACB = (180 - ∠ BAC) / 2 = (180-30) / 2 = 75 ? in quadrilateral BCPA, B, C, P, a four points are circular ? APC + ABC = 180 ? APC = 18