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Based on the above diagrams, the following relations hold true.
m∠ 1 &= 1/2(m BC - m AC) [0.25cm] m∠ 2 &= 1/2(m DEF - m DF) [0.25cm] m∠ 3 &= 1/2(m HI - m GJ)
Consider the first diagram and draw the auxiliary chord BC and point P.
By the Inscribed Angle Theorem, the measure of ∠ ABC is half the measure of AC. Also, the Tangent and Intersected Chord Theorem tells that the measure of ∠ PCB is half the measure of BC. m∠ ABC &= 1/2m AC [0.2cm] m∠ PCB &= 1/2m BC Next, the Triangle Exterior Angle Theorem says that m∠ PCB equals the sum of the measures of ∠ ABC and ∠ 1. Express the measure of ∠ 1 using this information. m∠ PCB = m∠ ABC + m∠ 1 ⇓ m∠ 1 = m∠ PCB - m∠ ABC Next, substitute the two angle measures found before into the right-hand side. m∠ 1 = m∠ PCB - m∠ ABC ⇓ m∠ 1 = 1/2m BC - 1/2m AC Finally, the proof can be completed by factoring 12 out.
m∠ 1 = 1/2(mBC - mAC)