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Based on the diagram above, the following relations hold true.
m∠ 1 &= 1/2(m AB + m CD) [0.25cm] m∠ 2 &= 1/2(m AD + m BC)
Let E be the point of intersection of chords AC and BD. Start by drawing an auxiliary chord BC.
Notice that ∠ 1 is an exterior angle of △ BCE. Therefore, by the Triangle Exterior Angle Theorem, its measure equals the sum of the measures of the two non-adjacent interior angles. m∠ 1 = m∠ ACB + m∠ CBD By the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of the intercepted arc. m∠ ACB = 12m AB & (I) m∠ CBD = 12m CD & (II) Finally, substituting these two equations into the equation given by the Triangle Exterior Angle Theorem, the first required equation is obtained.
m∠ 1 = 1/2m AB + 1/2m CD ⇓ m∠ 1 = 1/2(m AB + m CD)
To obtain the second equation, draw the auxiliary chord AB.
As before, ∠ 2 is an exterior angle of △ ABE. Therefore, its measure is equal to the sum of the measures of the two non-adjacent interior angles. m∠ 2 = m∠ ABD + m∠ CAB Once more, the Inscribed Angle Theorem can be applied to rewrite the two angle measures on the right-hand side in terms of the corresponding intercepted arcs. m∠ ABD = 12m AD & (I) m∠ CAB = 12m BC & (II) Finally, substitute the last two equations into the one relating m∠ 2 and the measure of the inscribed angles.
m∠ 2 = m∠ ABD + m∠ CAB ⇓ m∠ 2 = 1/2(m AD + m BC)