∴ arc AC= arc AB

∴∠AOB=2∠D

∫∠AOB = 50

∴∠D=25

∫∠AOC = 2∠d。

∴∠AOC=50

1.

Proof: O is OE, so that OE⊥AB and the vertical foot is E.

∴CE=DE,AE=BE

∴AE-CE=BE-DE

Namely AC=BD

1.

Solution: Let O point be OE, let OE⊥AB, let the vertical foot be E, extend the intersection CD of EO at F point, and connect OA and OD, using Pythagorean theorem: OE=

=

=5

Similar = 12

∴EF=OE+OF= 17 or EF=OF-OE=7.