As shown in the figure, it is known that ⊙ o is the circumscribed circle of ⊙ ABC, CE is the diameter of ⊥ o, CD ⊥ AB and D are perpendicular feet. It is proved that ⊙ ACD = ∠ BCE

As shown in the figure, it is known that ⊙ o is the circumscribed circle of ⊙ ABC, CE is the diameter of ⊥ o, CD ⊥ AB and D are perpendicular feet. It is proved that ⊙ ACD = ∠ BCE

Proof: connect EB,
∵CD⊥AB，
∴∠ADC=90°，
∴∠A+∠ACD=90°，
∵ CE is the diameter of ⊙ o,
∴∠CBE=90°，
∴∠E+∠ECB=90°，
∵∠A=∠E，
∴∠ACD=∠BCE．

A. B, C are on the circle O, CE is the diameter, CD is perpendicular to point D. It is proved that ∠ ACD = ∠ BCE

The vertical diameter theorem must be proved
We know that ad = BD, there is a vertical, there is a common side, so congruence, can be proved

As shown in the figure, it is known that ⊙ o is the circumscribed circle of ⊙ ABC, CE is the diameter of ⊥ o, CD ⊥ AB and D are perpendicular feet. It is proved that ⊙ ACD = ∠ BCE

Proof: connect EB,
∵CD⊥AB，
∴∠ADC=90°，
∴∠A+∠ACD=90°，
∵ CE is the diameter of ⊙ o,
∴∠CBE=90°，
∴∠E+∠ECB=90°，
∵∠A=∠E，
∴∠ACD=∠BCE．

As shown in the figure, AB parallel CD, AE bisect angle BAC, CE bisect angle ACD, and calculate the degree of angle E

The angle AEC is 90 degrees, because AB is parallel to CD, so the angle BAC plus angle ACD equals 180 degrees, and because AE bisects angle BAC, CE bisects angle ACD, so angle BAC equals 2 times angle EAC, angle ACD equals 2 times angle ace, so angle BAE plus angle EAC plus angle ace plus angle ECD equals 180 degrees, that is, 2 times angle EAC plus 2 times angle ECA equals 180 degrees

As shown in the figure AB parallel CD, ∠ AEC = 90 ° CE bisection ∠ ACD, it is proved that AE bisection ∠ BAC

It is proved that because AB is parallel to CD, so ∠ BAC + ∠ ACD = 180 degrees
And because ∠ AEC = 90, so ∠ ACE + ∠ CAE = 90
So ∠ BAC + ∠ ACD = ∠ BAE + ∠ CAE + ∠ ACE + ∠ DCE
Namely: ∠ BAE + ∠ DCE = 90
Because ∠ ace = ∠ DCE, ∠ BAE + ∠ ace = 90
So ∠ ace = ∠ BAE
That is, AE bisection ∠ BAC

Known, as shown in the figure, ab ∥ CD, AE bisection ∠ BAC, CE bisection ∠ ACD, verification: AE ⊥ CE

Proof: ∵ ab ∥ CD,
∴∠BAC+∠ACD=180°，
∵ AE bisection ∠ BAC, CE bisection  ACD,
∴∠EAC=1
2∠BAC，∠ACE=1
2∠ACD，
∴∠EAC+∠ACE=1
2（∠BAC+∠ACD）=90°，
∴∠AEC=180°-（∠EAC+∠ACE）=90°，
∴AE⊥CE．

As shown in the figure, AE and CE are bisectors of angle BAC and angle ACD respectively. If angle 1 + angle 2 = 90 degrees, is ab parallel to CD? Write the reason

Because AE CE bisects ∠ BAC and ∠ ACD
So ∠ 1 = ∠ BAE, ∠ 2 = ∠ DCE
Because it is 1 + ∠
So ∠ BAC + ∠ DCA = 180 °
So AB / / CD

As shown in the figure, AE and CE are bisectors of ∠ BAC and ∠ ACD respectively, and ∠ 1 + ∠ 2 = ∠ AEC (1) Try to determine the position relationship of line AB and CD; (2) Are lines AE and CE perpendicular to each other? If they are perpendicular to each other, please give proof; if not, give reasons

(1) As ∠ AEF = ∠ 1, EF intersects AC with F, as shown in Fig
∵∠BAE=∠1，
∴∠BAE=∠AEF，
∴AB∥EF．
∵∠1+∠2=∠AEC，
∴∠FEC=∠2．
And ∵ DCE = ∵ 2,
∴∠FEC=∠DCE，
∴CD∥EF，
∴AB∥CD．
（2）AE⊥CE．
∵AB∥CD，
∴∠BAC+∠ACD=180°．
∵∠BAC=2∠1，∠ACD=2∠2，
∴2∠1+2∠2=180°，
∴∠1+∠2=90°．
∵∠AEC=∠1+∠2，
∴∠AEC=90°，
∴AE⊥CE．

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, AB is the diameter of ⊙ o, P is a point on the extension line of AB, PD cuts ⊙ o at point C, the extension lines of BC and ad intersect at point E, and ad ⊥ PD (1) Results: ab = AE; (2) When the value of AB: BP is, the △ Abe is an equilateral triangle and the reason is given

(1) It is proved that: connect OC, ∵ PD cut ⊙ o at point C,