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{{ printedBook.courseTrack.name }} {{ printedBook.name }} 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.

$m∠1=21 mAC$ and $m∠2=21 mADC$

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, $m∠CAF=21 mFC$ 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.

$m∠1=21 mAC$

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.

$m∠2=21 mADC$