1. As shown in the figure, a is the point on ⊙ o, the extension line of tangent intersection diameter CB passing through point a is at point P, ad ⊥ BC is at point D 1. As shown in the figure, a is the point on ⊙ o, the extension line of tangent diameter CB passing through point a is at point P, ad ⊥ BC is at point D Verification: Pb / PD = PO / PC Indicate BDO three points collinear

1. As shown in the figure, a is the point on ⊙ o, the extension line of tangent intersection diameter CB passing through point a is at point P, ad ⊥ BC is at point D 1. As shown in the figure, a is the point on ⊙ o, the extension line of tangent diameter CB passing through point a is at point P, ad ⊥ BC is at point D Verification: Pb / PD = PO / PC Indicate BDO three points collinear

According to the tangent theorem, we can know that PA ^ 2 = Pb * PC
Right triangle pad is similar to right triangle POA, so their corresponding edges are also similar
So PO / PA = PA / PD, PA ^ 2 = Po * PD
It is concluded that Po * PD = PA ^ 2 = Pb * PC
So Pb / PD = PO / PC

Circle OA and circle ob circumscribe at point P, BC tangent circle OA and point C, inner common tangent PD of circle OA and circle ob intersect AC and point D, intersect BC and m, find CD = Pb

∵ BC is the tangent line of circle a, PD is the tangent line of circle a ∵ AC ⊥ BC (AD ⊥ BC), that is, ∵ DCM = 90 ° (the radius of AC is perpendicular to the tangent line) MC = MP (two tangents from a point outside the circle to the circle are equal) ∵ PD is the tangent line of circle B ∵ BP ⊥ PD, that is, ∵ BPM = ∵ DCM = 90 ° in △ CDM and △ PBM ∵ BMP = ∵ DMC (opposite vertex) ∵ BPM = ∵ DC

It is known that AB and CD intersect at O, ad and CB, and the long line intersects at e, OA = OC, EA = EC

I'm afraid you don't understand. Connecting AC in the triangle EAC, because EA = EC, the angle EAC = angle ECA is the same as the angle OAC = angle OCA, so the angle OAE = angle OCE is obtained, and then the angle a is equal to the angle c is obtained by the complementary angle relationship