Use the information given in the diagram to prove that m∠JGI = One-half(b – a), where a and b represent the degree measures of a
rcs FH and JI. A circle is shown. Secants G J and G I intersect at point G outside of the circle. Secant G J intersects the circle at point F. Secant G I intersects the circle at point H. The measure of arc F H is a. The measure of arc J I is b. A dotted line is drawn from point J to point H. Angles JHI and GJH are inscribed angles. We have that m∠JHI = One-half b and m∠GJH = One-halfa by the . Angle JHI is an exterior angle of triangle . Because the measure of an exterior angle is equal to the sum of the measures of the remote interior angles, m∠JHI = m∠JGI + m∠GJH. By the , One-halfb = m∠JGI + One-halfa. Using the subtraction property, m∠JGI = One-halfb – One-halfa. Therefore, m∠JGI = One-half(b – a) by the distributive property.
Angles JHI and GJH are inscribed angles. We have that m∠JHI = One-half b and m∠GJH = One-halfa by the <u>Inscribed angle theorem</u> . Angle JHI is an exterior angle of triangle <u>GJH </u>. Because the measure of an exterior angle is equal to the sum of the measures of the remote interior angles, m∠JHI = m∠JGI + m∠GJH. By the <u>Substitution property</u> , One-halfb = m∠JGI + One-halfa. Using the subtraction property, m∠JGI = One-halfb – One-halfa. Therefore, m∠JGI = One-half(b – a) by the distributive property.
