As shown in the figure, it is known that AB is the diameter of ⊙ o, Pb is the tangent of ⊙ o, B is the tangent point, Op ⊥ chord BC is at point D and intersects ⊙ o at point E. (1) prove: ∠ OPB = ∠ AEC; (2) if point C is the triad point of semicircle and ACB, please judge which special quadrilateral AOEC is? And explain the reason

As shown in the figure, it is known that AB is the diameter of ⊙ o, Pb is the tangent of ⊙ o, B is the tangent point, Op ⊥ chord BC is at point D and intersects ⊙ o at point E. (1) prove: ∠ OPB = ∠ AEC; (2) if point C is the triad point of semicircle and ACB, please judge which special quadrilateral AOEC is? And explain the reason

(1) It is proved that: ∵ AB is the diameter of ⊙ o, Pb is the tangent of ⊙ o, ∵ Pb ⊥ ab. ∵ OPB + ∠ POB = 90 °. ∵ op ⊥ BC, ∵ ABC + ∠ POB = 90 °. ∵ ABC = ∠ OPB. Also ∵ AEC = ∠ ABC, ∵ OPB = ∠ AEC. (2) the quadrilateral AOEC is rhombic

As shown in the figure, AB is the diameter of circle O, BD and PD tangent circle O at points B and C, P, a and B are collinear, and prove that Po × Pb = PC × PD

Certification:
∵ BD and PD are tangent lines of circle o
∴∠PCO=∠PBD=90º
And ∵ ∠ OPC = ∠ DPB [common angle]
∴⊿OPC∽⊿DPB（AA’）
∴PO/PD=PC/PB
∴PO×PB=PC×PD

PA.PB Tangent circle O to two points a and B, tangent through P, is called circle to C. D, parallel to CD through B, connect AE and PD to m, and prove that M is the midpoint of DC
Urgent solution! Thank you very much!

Let's do it! AB intersects CD with n ﹥ PAB = ﹥ BAE = ﹥ PMA, so ⊿ pan is similar to ⊿ PMA, so PC * PD = PA * PA = PN * PM. It is proved that P, C, N.D harmonic point sequence (PC / CN = Pd / DN) PC / PD = (PC / DN) * (dN / CN) = (AC / AD) * (CB / BD) (left and right triangles are similar) = (AC * CB * sin ﹥ ACB) / (ad * BD * sin