a)

CDB = CAB (angles in the same segment CB)

KDO = KAO (angles in the same segment KO of the small circle AOKD

KDO = ODB (common angle, BKD is straight line and OD is common)

CAB = KAO (common angle, AKC and AOB are straight lines)

So ODB = KDO = KAO = CAB = CDB ①


OBD = ODB (base angles of isosceles triangle ODB, OD and OB are both radius of the circle ABCD)

ACD = ABD (angles in the same segment AD)

Now ABD = OBD (common angle, AOB and BKD are straight lines)

So ACD = ABD = OBD = ODB ②

From ① and ② we can conclude that :

ACD = ABD = OBD = ODB = KDO = KAO = CAB = CDB

Therefore ACD = CDB .


Now,
ACD = KCD (common angle, AKC is straight line and CD is common)

And CDK = CDB (common angle , BKD is a straight line and CD is common)

So KCD = ACD = CDB = CDK

Therefore KCD = CDK. Which means triangle KCD is isosceles.

So DK = CK (legs of isosceles triangle are equal, DC is the base)

b)

From a) we know that ODB = CDB. And since ODB + CDB = ODC, BKD must be the angle bisector of ODC