OA and OB are radii, so the triangle OAB is an isosceles triangle, so the angle OBA=20 degrees and the angle AOB= 140 degrees; Because A and B are the tangent points of a circle, the angle OAP= the angle OBP=90 degrees, and OAPB is a quadrilateral, so the internal angle is 360 degrees, and the angle P=360-90-90- 140=40 degrees.
Question 2: (1) Proof:
From the meaning of the question, we can see that the triangle CED and the triangle CAB are isosceles triangles, and the angles CDE and CBA correspond to the arc AC, so the two angles are equal. It can be concluded that triangle CED is similar to triangle CAB, so angle ECA+ angle ACD= angle ACD+ angle DCB, so angle ECA= angle DCB, and because EC=DC, AC=BC, angle ECA= angle.
(2) Because AC is perpendicular to BC and AC=BC, the triangle ABC is an isosceles right triangle, and because the triangle CED is similar to the triangle CAB, the triangle CED is an isosceles right triangle, so ED= root 2CD, and because ED=EA+AD, EA=BD, BD+AD= root 2CD.
Question 3: (1) According to the meaning of the question, the triangle BAP is a right triangle, so AB/AP=tan30= (root number 3)/3, then AP=AB/[ (root number 3)/3]=2 root number 3.
(2) Connect Co, CA, OB=OC, and the angle ABD=60, so the angle BAC=30, and because the angle BAD=90, so the angle CAD=60, and because the angle P=30 and D is the midpoint of AP, so the triangle ACD is an isosceles triangle, so the angle BCO= the angle ACD=60, and because the angle BCO+.