Known, as shown in the figure, ab ‖ CD, AE bisection ∠ BAC, CE bisection ∠ ACD, find the degree of ∠ E

Known, as shown in the figure, ab ‖ CD, AE bisection ∠ BAC, CE bisection ∠ ACD, find the degree of ∠ E

∵ ab ∥ CD, AE bisection ∵ BAC, CE bisection ∠ ACD,
And ∠ BAC + ∠ DCA = 180 ° {CAE + ∠ ace = 1
2（∠BAC+∠DCA）=90°，
∠E=180°-（∠CAE+∠ACE）=90°，
∴∠E=90°．

As shown in the figure, if be bisection ∠ ABC, CE bisection ∠ BCD, and ∠ 1 + 10 ∠ 2 = 90 °, is ab ‖ CD? Why?

AB∥CD，
Reason: ∵ be bisection ∠ ABC, CE bisection ∠ BCD,
∴∠1=∠ABE，∠2=∠ECD，
∵ 1 × 10  2 = 90 °,
∴∠ABE+∠1+∠2+∠ECD=180°，
That is ∠ ABC + ∠ BCD = 180 °,
∴AB∥CD．

As shown in the figure, be bisection ∠ ABC, CE bisection ∠ ACD is known, and be is given to E. verification: AE bisection ∠ fac

It is proved that: as shown in the figure, the crossing point E is eg ⊥ BD, eh ⊥ Ba, EI ⊥ AC, and the vertical feet are g, h and I respectively,
∵ be bisection ∵ ABC, eg ⊥ BD, eh ⊥ Ba,
∴EH=EG．
∵ CE bisection ∵ ACD, eg ⊥ BD, EI ⊥ AC,
∴EI=EG，
Ψ EI = eh (equivalent substitution),
Ψ AE bisection ∠ fac (points with equal distance to both sides of the corner must be on the bisector of the angle)

As shown in the figure, be bisection ∠ ABC, CE bisection ∠ ACD is known, and be is given to E. verification: AE bisection ∠ fac

It is proved that: as shown in the figure, the crossing point E is eg ⊥ BD, eh ⊥ Ba, EI ⊥ AC, and the vertical feet are g, h and I respectively,
∵ be bisection ∵ ABC, eg ⊥ BD, eh ⊥ Ba,
∴EH=EG．
∵ CE bisection ∵ ACD, eg ⊥ BD, EI ⊥ AC,
∴EI=EG，
Ψ EI = eh (equivalent substitution),
Ψ AE bisection ∠ fac (points with equal distance to both sides of the corner must be on the bisector of the angle)

As shown in the figure, be bisection ∠ ABC, CE bisection ∠ ACD is known, and be is given to E. verification: AE bisection ∠ fac

It is proved that: as shown in the figure, the crossing point E is eg ⊥ BD, eh ⊥ Ba, EI ⊥ AC, and the vertical feet are g, h and I respectively,
∵ be bisection ∵ ABC, eg ⊥ BD, eh ⊥ Ba,
∴EH=EG．
∵ CE bisection ∵ ACD, eg ⊥ BD, EI ⊥ AC,
∴EI=EG，
Ψ EI = eh (equivalent substitution),
Ψ AE bisection ∠ fac (points with equal distance to both sides of the corner must be on the bisector of the angle)

As shown in the figure, be bisection ∠ ABC, CE bisection ∠ ACD is known, and be is given to E. verification: AE bisection ∠ fac

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prove:
Connect OP, because P is the tangent point of line CD and circle o
So: OP is perpendicular to CD
So: AD / / op / / BC
Because AB is the diameter, OA = ob
So PD = PC

As shown in the figure, AB is the diameter of ⊙ o, point P is on the extension line of Ba, chord CD ⊥ AB is at point E, ∠ POC = ∠ PCE (1) Verification: PC is tangent of ⊙ o (2) If OE: EA = 1:2, PA = 6, find the radius of ⊙ o

It is proved that: (1) ∵ CD ⊥ AB,  CEO = 90 °, ∵ PCE = ∵ POC,  PCE + ∠ OCD = 90 °, OC ⊥ PC, and ∵ OC is the radius, ? PC is the tangent of ⊙ o; (2) let OE = k, then AE = 2K, OC = 3k, in RT △ OCE, CE = 22K, ? P =

It is known that: as shown in the figure, AB is the diameter of ⊙ o, point P is on the extension line of Ba, PD cuts ⊙ o at point C, BD ⊥ PD, the perpendicular foot is D, connecting BC The results are as follows: (1) BC bisection ∠ PBD; （2）BC2=AB•BD．

Proof: (1) connection OC. (1 point)
∵ PD tangent ⊙ o at point C,
And ∵ BD ⊥ PD,
∴OC∥BD．
Ψ 1 = ∠ 3. (2 points)
And ∵ OC = ob,
Ψ 2 = ∠ 3. (3 points)
Ψ 1 = ∠ 2, i.e. BC bisection ∠ PBD. (4 points)
(2) Connect AC
∵ AB is the diameter of ⊙ o,
﹤ ACB = 90 °. (5 points)
And ∵ BD ⊥ PD,
Ψ ACB = ∠ CDB = 90 ° (6 points)
And ∵ 1 = ∠ 2,
﹤ ABC ∽ CBD; (7 points)
∴AB
CB＝BC
BD, ν BC2 = ab · BD. (8 points)

As shown in the figure, ad is the diameter of circle O, BC edge of △ ABCD passes through point D, AB, AC and circle O intersect at points E and F, cut AE * AB = af * AC, verify; BC is tangent line of circle o As shown in the figure, ad is the diameter of circle O, BC is the tangent line, the tangent point is point D, and AB, AC and circle O intersect at points E and F, and verify; AE * AB = af * AC

Ad is the diameter of circle O, BC edge of △ ABCD passes through point D, AB, AC and circle O intersect at points E and F, cut AE * AB = af * AC, verify; BC is tangent line of circle o
prove:
∵AE*AB=AF*AC
∴AE/AC=AF/AB,
Angle BAC = angle cab
So △ AEF ∽ ABC
Angle AEF = angle c
The angle AEF and angle ADF are the circular angles of the same arc, so the angle AEF = angle ADF
Then the angle ADF = angle C
The angle ADF + angle DAF = 90, so the angle c + angle DAF = 90
So ad ⊥ BC, and ad is the diameter of circle O, so BC is the tangent of circle o