Question

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 arcs 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
JIH
. 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.

Answer 1

Fill in the missing reasons and complete the proof:

m∠JHI = 1/2 b and m∠GJH = 1/2 a by the Inscribed Angle Theorem (an inscribed angle measures half its intercepted arc).
Angle JHI is an exterior angle of triangle JIH. By the Exterior Angle Theorem, m∠JHI = m∠JGI + m∠GJH.
Substitute the inscribed-angle values: 1/2 b = m∠JGI + 1/2 a (by Substitution).
Subtract 1/2 a from both sides: m∠JGI = 1/2 b − 1/2 a.
Factor 1/2: m∠JGI = 1/2 (b − a).

Thus m∠JGI = one-half(b − a).