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
. 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 argument:

1. m∠JHI = 1/2 b and m∠GJH = 1/2 a by the Inscribed Angle Theorem (each inscribed angle measures half its intercepted arc).

2. Angle JHI is an exterior angle of triangle JGH (HI is the extension of HG).

3. By the Exterior Angle Theorem, the exterior angle equals the sum of the two remote interior angles:
m∠JHI = m∠JGI + m∠GJH.

4. Substitute the inscribed-angle values: 1/2 b = m∠JGI + 1/2 a.

5. Rearranging gives m∠JGI = 1/2 b − 1/2 a = 1/2 (b − a) (by factoring 1/2).

Therefore m∠JGI = 1/2 (b − a).