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: mABC = mCBV.

 

[Problem submitted by Roger Wolf, Chairman of the Math Department, LACC.]

 

 

Solution for Problem 9:

 

mOTV=90 &  mOAT=90 OTA~OVT   

OT=OB    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