**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 = m_{}CBV.

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

_{}

**Solution for Problem 9:**

m_{}OTV=90_{} & m_{}OAT=90_{} _{}_{}OTA~_{}OVT _{} _{}

OT=_{} _{} _{} _{}OBA ~ _{}OVB _{} m_{}OBA =m_{}OVB

m_{}OVB=m _{}CVB _{} m_{}OBA=m _{}CVB--------------(1)

m_{}OBC=m_{}OBA+ m_{}ABC

m_{}OCB= m_{}CBV+ m_{}CVB

_{}m_{}OBC= m_{}OCB _{} m_{}OBA+ m_{}ABC= m_{}CBV+ m_{}CVB----(2)

By
equations (1) and (2), m_{}ABC= m_{}CBV