Om? Open? OP is a vertical line obtained by connecting tangent points, which can prove that triangle OBN is congruent with triangle OBM, and OCN is congruent with OCP.

Extend the intersection of AB and DC at point Q and connect OQ, and the ratio QOM=QOP= angle A/D, angle NQO=CQO= angle POD=MOA is obtained.

In a word, Jiaoka =POQ=QOB (angle OMQ = OPQ = 90°).

So the angle POC=QOB and the angle OQB=DOP, then add the above two diagonal angles, and the angle ABO=COD, then the triangle AOB is similar to the triangle DCO, and the answer is acceptable.

(The picture is a little rough, please forgive me. . . )