If a tangent and a chord intersect at a point on a circle, then the measure of each angle formed is one-half the measure of its intercepted arc.
Based on the diagram, the following relation holds.
m∠1=21mAC and m∠2=21mADC
Keep in mind that this theorem is true also in the case that AB is a secant.
Consider a diameter AF. Since AB is tangent to the circle at A, then AF⊥AB.
In addition, since AF is a diameter, ACF is a semicircle and then its measure is 180∘. The Arc Addition Postulate allows setting the following equation. mAC+mFC=180∘mACF Since ∠CAF is a central angle, its measure is half the measure of the intercepted arc. That is, m∠CAF=21mFC or equivalently, mFC=2m∠CAF. mAC+2m∠CAF=180∘ Similarly, the Angle Addition Postulate can be used to establish the following equation. m∠1+m∠CAF=90∘ Next, multiply the equation above by -2 and add it to the previous equation. mAC+2m∠CAF+-2m∠1−2m∠CAFmAC−2m∠1=180∘=-180∘=0 Solving the last equation by m∠1 gives the first desired equation.
m∠1=21mAC
Once more, the Arc Addition Postulate and the Angle Addition Postulate can be applied to set the following pair of equations. {mADF+mFC=mADCm∠EAF+m∠CAF=m∠2(I)(II) Recall that mADF=180∘, m∠EAF=90∘, and mFC=2m∠CAF. {180∘+2m∠CAF=mADC90∘+m∠CAF=m∠2(I)(II) Next, multiply Equation (II) by -2 and add it to Equation (I). 180∘+2m∠CAF+-180∘−2m∠CAF0=mADC=-2m∠2=mADC−2m∠2 Finally, by solving the resulting equation for m∠2, the second desired equation is obtained.
m∠2=21mADC