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: mABC = mCBV.
[Problem submitted by Roger Wolf, Chairman of the Math Department, LACC.]
Solution for Problem 9:
mOTV=90 & mOAT=90 OTA~OVT
mOVB=m CVB mOBA=m CVB--------------(1)
mOCB= mCBV+ mCVB
mOBC= mOCB mOBA+ mABC= mCBV+ mCVB----(2)
By equations (1) and (2), mABC= mCBV