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.

See the attached figure which represents the problem.

Angles GJH and JHI are inscribed angles

Given: ∠GJH = 0.5a  and ∠JHI = 0.5b ⇒ inscribed angle theorem

So, the angle JHI is an exterior angle of ΔGJH

AS, the measure of the exterior angle is equal to the sum of the sum of the remote interior angles

So, ∠JHI = ∠GJH + ∠JGI ⇒ by substitution property

∴ 0.5 b = 0.5a + ∠JGI

∴ ∠JGI = 0.5b - 0.5a ⇒ take 0.5 as a common

∠JGI = 0.5 ( b - a ) ⇒ by distributive property.

So, m∠JGI = One-half(b – a)

see the explanation

Step-by-step explanation:

The picture of the question in the attached figure

we know that

The measure of the exterior angle is the semi-difference of the arches it covers.

In this problem

m∠JGI is an exterior angle

so

we have

substitute the given values

Prove

Remember that

In any triangle the measure of an exterior angle is equal to the sum of the measures of the remote interior angles

so

In the triangle GJH

----> equation A

Remember that

The inscribed angle is half that of the arc it comprises.

so

substitute in the equation A

Using the subtraction property

simplify

----> proved

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step-by-step explanation:

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