As shown in Fig. 1, ⊙ o is the circumscribed circle of ⊙ ABC, and ∠ B = ∠ CAD

prove:
[the center angle corresponding to the same arc is 2 times of the circumference angle]
So ∠ AOC = 2 ∠ B
In triangle AOC, ∠ OAC = (180 - ∠ AOC) / 2 = 90 - ∠ AOC / 2 = 90 - ∠ B
So ∠ OAC + ∠ B = 90 °
Because of ∠ B = ∠ CAD
So ∠ OAC + ∠ CAD = 90 °
OA ⊥ ad
So ad is tangent to ⊙ o
Conclusion

As shown in the figure, △ ABC is the inscribed triangle of ⊙ o, and ad is the diameter of ⊙ O. if ∠ ABC = 50 °, calculate the degree of ⊙ CAD

Connect the CD;
Then ∠ ADC = ∠ ABC = 50 °
∵ ad is the diameter of ⊙ o,
∴∠ACD=90°
∴∠CAD+∠ADC=90°
∴∠CAD=90°-∠ADC=90°-50°=40°．

It is known that the triangle ABC is inscribed on the circle O, ∠ CAD = ∠ ABC, and the position relationship between the straight line AD and the circle O can be judged

The line ad is tangent to the circle o
Prove: connect AO and extend intersection circle O to e, connect CE
If AE is the diameter, then: ∠ ace = 90 °, CAE + e = 90 °
∵∠E=∠ABC;∠CAD=∠ABC.
Since ∠ CAD = ∠ e, so ∠ CAE + ∠ CAD = 90 °, then ad is tangent to circle o

Circle O is the circumscribed circle of triangle ABC, BC is the diameter of circle O. extend BC to D and connect ad so that ∠ CDA + 2 ∠ CAD = 90 ° solution: ad is the tangent line of circle o

According to the triangle exterior angle theorem, there are: ∠ OCA = ∠ CDA ＋∠ CAD
Obviously, there are: OA = OC, ﹤ OAC = ∠ OCA, ﹥ OAC = ∠ CDA ﹣ CAD,
 OAC ⊙ CAD = CDA + 2 ∠ CAD = 90 °, oad = 90 °, ad cut ⊙ o to a

In the triangle ABC, the angle a = 60 degrees, BC is the fixed length, and the circle O with BC as its diameter intersects AB and AC at points D and e respectively to connect de and OE

The following conclusions: 1, BC = 2DE; 2, D to OE distance unchanged; 3, BD + CE = 2DE; 4, OE is the tangent of the triangle ade circumscribed circle

In the triangle ABC, the angle B = 60 ° and the bisector ad, CE intersect with O. it is proved that AE + CD = AC

Take a point on AC and connect of so that angle AOF = angle AOE
Angle AOE = 180 degrees - angle bad - angle AEO
=180 degree - (1 / 2) angle BAC - (angle ABC + angle BCE)
=180 degrees - (1 / 2) (angle BAC + angle ACB) - angle ABC
=180 degrees - (1 / 2) (180 degrees angle ABC) - angle ABC
=90 degree - (1 / 2) angle ABC
=60 degrees
ν: angle AOF = angle AOE = angle doc = 60 degrees
Angle COF = 180 degrees - angle AOE - angle AOF = 60 degrees = angle doc
Triangle AEO is equal to triangle AFO
Triangle doc is equal to triangle FOC
AE=AF
DC=FC
∴AE+CD=AF+FC=AC

As shown in the figure, △ ABC and △ ade are isosceles right triangles. CE and BD intersect at point m and BD intersect AC at point n It is proved that: (1) BD = CE; (2) BD ⊥ CE

It is proved that: (1) ∵ BAC = ∵ DAE = 90 
∴∠BAC+∠CAD=∠DAE+∠CAD
That is ∠ CAE = ∠ bad
In △ abd and △ ace
AB＝AC
∠CAE＝∠BAD
AD＝AE
∴△ABD≌△ACE（SAS）
∴BD=CE
（2）∵△ABD≌△ACE
∴∠ABN=∠ACE
∵∠ANB=∠CND
∴∠ABN+∠ANB=∠CND+∠NCE=90°
∴∠CMN=90°
BD ⊥ CE

In the triangle ABC, the angle BAC is equal to 90 degrees, AB is equal to AC, AE is the straight line passing through point a, BD is perpendicular to point E, and it is proved that BD is equal to CE plus De

From the condition: ab = AC, (1)
∠ABD+∠BAD=90°,
∠CAE+∠BAD=90°,
∴∠ABD=∠CAE（2）
And △ abd and △ CAE are right triangles,
According to conditions (1) and (2)
△ABD≌△CAE（A,A,S）,
∴BD=AE,
Ad = CE,
∴AE=AD+DE=CE+DE,
∴BD=CE+DE.
The proof is over

In RT △ ABC, ∠ a = 90 ° CE is an angular bisector, which intersects with high ad at F, and FG ‖ BC intersects AB with G, Confirmation: AE = BG

Make eh ⊥ BC in H, as shown in the figure, ∵ e is the point on the bisector, eh ⊥ BC, EA ⊥ Ca, ᙽ EA = eh, ∵ ad is the height of ⊥ ABC, EC bisection ∠ ACD, ∠ ADC = 90 °, ACE = ∠ ECB, ∵ B = ∠ DAC, ? AEC = ∠ DAC + ∠ ECA = ∠ AE = AF, ﹤ eh = AF

As shown in Fig. 1, ⊙ o is the circumscribed circle of ⊙ ABC, and ∠ B = ∠ CAD