Problem 9
Given: the center O of the small
circle is on the large circle. A diameter OV of the large circle intersects the
common cord TU of the circles at A, and the small circle at C. B is a point on
the small circle as shown. Prove: m
ABC = mCBV.
[Problem submitted by Roger
Wolf, Chairman of the Math Department, LACC.]
Solution for Problem 9:
mOTV=90
& mOAT=90 
OTA~OVT 
OT= 
OBA ~ OVB mOBA =mOVB
mOVB=m CVB mOBA=m CVB--------------(1)
mOBC=mOBA+ mABC
mOCB= mCBV+ mCVB
mOBC= mOCB mOBA+ mABC= mCBV+ mCVB----(2)
By
equations (1) and (2), mABC= mCBV