By the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of its intercepted arc.
m∠ECD = 12m DE & (I) m∠BDC = 12m BC & (II)
Notice that ∠ECD is an exterior angle of △ ACD. Then, applying the Triangle Exterior Angle Theorem, we conclude that its measure is equal to the sum of the measures of the two nonadjacent interior angles.
m∠ECD = m∠BDC + m∠A
⇓
m∠A = m∠ECD - m∠BDC
Finally, we substitute the two equations found above into the last equation. That way, we get the desired equation.
m∠A = 1/2m DE - 1/2m BC
⇓
m∠A = 1/2(m DE - m BC)
Two-Column Proof
Given: & SecantsADandAE
Prove: & m∠A = 12(mDE - mBC)
Let's summarize the proof we did above in the following two-column proof table.
