As shown in the figure, in the two concentric circles with point o as the center, the chord ab of the large circle is the tangent line of the small circle, and the point P is the tangent point

As shown in the figure, in the two concentric circles with point o as the center, the chord ab of the large circle is the tangent line of the small circle, and the point P is the tangent point

Proof: connect OP as shown in the figure,
∵ the chord ab of the big circle is the tangent line of the small circle, and the point P is the tangent point,
∴OP⊥AB，
∵ OP over O,
∴AP=BP．

As shown in the figure, in the two concentric circles with point o as the center, the chord ab of the large circle is the tangent line of the small circle, and the point P is the tangent point

Proof: connect OP as shown in the figure,
∵ the chord ab of the big circle is the tangent line of the small circle, and the point P is the tangent point,
∴OP⊥AB，
∵ OP over O,
∴AP=BP．

As shown in the figure, in the two concentric circles with point o as the center, the chord ab of the large circle is the tangent line of the small circle, and the point P is the tangent point

Proof: connect OP as shown in the figure,
∵ the chord ab of the big circle is the tangent line of the small circle, and the point P is the tangent point,
∴OP⊥AB，
∵ OP over O,
∴AP=BP．

As shown in the figure, in the two concentric circles with point o as the center, the chord ab of the large circle is the tangent line of the small circle, and the point P is the tangent point

Proof: connect OP as shown in the figure,
∵ the chord ab of the big circle is the tangent line of the small circle, and the point P is the tangent point,
∴OP⊥AB，
∵ OP over O,
∴AP=BP．

As shown in the figure, AB is the diameter of circle O, and the chord CD intersects AB at point P ∠ APD = 45 °, AP = 5, Pb = 1 to find the length of CD

O for OE ⊥ CD for e
Connect OD
∵AP=5,PB=1
∴AB=5+1=6
∵ AB is the diameter of circle o
∴OD=OB=3
∵PB=1
∴OP=2
∵∠APD=45°
∴OE=√2
∴ED²=OD²-OE²
=9-2
=7
∴ED=√7
CD=2√7

As shown in the figure, AB is the diameter of circle O, chord CD intersects with ab at point P, the angle AOD is equal to 70 ° and the angle APD is equal to 60 ° to find the degree of angle BDC

∵ AOD is the center angle corresponding to the circumference angle  abd
∴∠ABD＝∠AOD/2＝70/2＝35
∵OB＝OD
∴∠BDO＝∠ABD＝35
∵∠AOD＝∠APD+∠CDO
∴∠CDO＝∠AOD-∠APD＝70-60＝10
∴∠BDC＝∠BDO-∠CDO＝35-10＝25°
The math group answered your call for help,

As shown in the figure, AB is the diameter of ⊙ o, the chord CD intersects AB at P, and ﹤ APD = 60 ° and ﹤ cob = 30 °______ .

∵∠APD=∠C+∠COB，
∴∠C=∠APD-∠COB=60°-30°=30°，
∴∠ABD=∠C=30°．
So the answer is 30 degrees

The chords AB and CD of circle O intersect at the bisection angle APD of P and Po, and find AB = CD

(when drawing, AB CD is on the same side of the center of the circle)
The perpendicular line of AB CD passing through o intersects with e F
Po bisects ∠ APD, i.e., ∠ ope = ∠ OPF
OP=OP,
There is RT △ ope ≌ RT △ OPF
Then OE = of
Then AB = 2V (R ^ 2-oe ^ 2) = 2V (R ^ 2-of ^ 2) = CD

In the circle O, the chords AB and CD intersect at P, and ab = CD. It is proved that Po bisects ∠ DPB

Tell you the simplest proof
E is the midpoint of AB and F is the midpoint of CD
Because AB = CD, AE = 1 / 2 ab; CF = 1 / 2 CD
Because OA = OC = R (circle radius), OE = of
So o is on the bisector of angle DPB, so Po bisection ∠ DPB
Conclusion
Ask me which step you can't understand

Let p be proved by the point of intersection of DC, P and B

Connect OA, ob, OC, OD, and make OE vertical AB through O. the intersection point is e, of vertical CD, and the intersection point is f
The angle OEP = angle OFP = 90 ° and Po bisects the angle DPB, and OP is the common edge
So the triangle OEP is all equal to the triangle ofP
So OE = of
Because ob = OD, a right angle side and an oblique side are equal
OBE of right triangle is equal to ODF of right triangle
So be = DF
In the isosceles triangle AOB and isosceles triangle cod
AB=2BE,DC=2DF
So AB = DC