As shown in the figure, BD is the diameter of ⊙ o, point a is the midpoint of arc BC, ad intersects BC at point E, AE = 2, ed = 4 (1) The results were as follows: △ Abe ∽ abd; (2) Find the value of Tan ∠ ADB; (3) Extend BC to f and connect FD so that the area of △ BDF is equal to 8 3. Find the degree of ∠ EDF

(1) It is proved that: ∵ point a is the middle point of arc BC,  ABC = ∠ ADB, and ? BAE = ∵ BAE, ∵ Abe ∵ ADB. (3 points) (2) ∵ ∵ Abe ∵ ADB, ∵ AB2 = 2 × 6 = 12, ? AB = 23. In RT △ ADB, Tan  ADB = 236 = 33. (3) (3) connect CD, then ? BCD = 90 °

As shown in the figure, a, B, C, D are the four points on ⊙ o, ab = AC, ad intersects BC at point E, AE = 3, ed = 4, then the length of AB is () A. 3 B. 2 Three C. Twenty-one D. 3 Five

∵AB=AC，
∴∠ACB=∠ABC=∠D，
∵∠BAD=∠BAD，
∴△ABD∽△AEB，
∴AB
AE＝AD
AB，
∴AB2=3×7=21，
∴AB=
21．
Therefore, C

As shown in the figure, △ ABC is inscribed in ⊙ o, ab = AC, chord ad intersects BC at point E, AE = 4, ed = 5, (1) Verification: ad bisection ∠ BDC; (2) Find the length of AC; (3) If the bisector Ci of BCD intersects with AD at point I, it is proved that AI = AC

0

∵ AB is the diameter,
∴∠ACB=90°，
∵∠CAB=30°，
Ψ ABC = 60 °, the degree of arc BC = 1
2 the degree of arc AC;
∵AD=DC，
The degree of arc ad = degree of arc DC = 1
2 the degree of arc AC, ν degree of arc BC = degree of arc ad;
∴BC=AD．
In RT △ ABC
∵∠CAB=30°，AC=2
3 and BC = AC · Tan ∠ cab,
∴BC=2
3×tan30°=2．
∴AD=2．

It is known that, as shown in the figure, AB is the diameter of semicircle o, point C is a point on the semicircle, and through the point C, CD ⊥ AB is in D, AC = 2 Ad: DB = 4:1, calculate the length of AD

Connect BC
∵ AB is the diameter of semicircle o,
∴∠ACB=90°．
∵CD⊥AB，
∴∠ADC=90°．
∴∠ACB=∠ADC．
∵∠A=∠A，
∴△ACD∽△ABC．
∴AC
AB＝AD
AC．
Let DB = xcm, then ad = 4xcm, ab = 5xcm
∴2
Ten
5x＝4x
Two
10．
That is, 5x × 4x = (2
10）2．
The solution is X=
2．
∴AD=4
2cm．

It is known that, as shown in the figure △ ABC, ∠ C = 90 ° ad bisection ∠ BAC, CD = root 3, BD = 2 times root 3, calculate the length of bisector AD and ab AC

AB is the diameter of circle O, CD is perpendicular to point D, AC is equal to 3bC is equal to 4, find CD, ad, DB

Because AB is the diameter, so the angle c is equal to 90 degrees, because AC = 4 BC = 4 / 3, so AB = 8 times root 2. According to the formula BC square = BD * AB, BD = 9 / 9 root 2, ad = 71 / 9 root 2, CD = 3 / 14

It is known that OA and ob are the radius of circle O, C and D are the points on OA and ob respectively, and AC = BD. it is proved that ad = BC

OA = ob, AC = BD, so oa-ac = ob-bd, so OC = OD,
∠AOB=∠BOC,
So △ AOD ≌ △ BOC (SAS),
So ad = BC

It is known that C D in the circle O is the midpoint of the radius OA ob, and the intersection of ad BC and P proves that PA = Pb

Because C D is the midpoint of the radius OA ob, CO = do and AO = Bo ∠ AOB = ∠ boa, so ⊿ AOD is equal to ⊿ BOC, so ⊿ a = ∠ B and AC = BD cut ﹣ APC = ∠ BPD, so ⊿ APC is equal to ⊿ BPD, so PA = PB

Like OA, ob is the radius of circle O, C and D are the midpoint of OA and ob respectively. If the length of ad is 3cm, calculate the length of BC

∵ OA = ob (radius)
OC = OD (half of radius)
Common to AOB
∴ △AOD ≌ △ BOC
∴ BC = AD = 3 cm