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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.

Respuesta :

Matheng

Answer:

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)

Ver imagen Matheng
calculista

Answer:

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

[tex]m\angle JGI=\frac{1}{2}[arc\ JI-arc\ FH][/tex]

we have

[tex]arc\ JI=b^o\\arc\ FH=a^o[/tex]

substitute the given values

[tex]m\angle JGI=\frac{1}{2}(b-a)^o[/tex]

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

[tex]m\angle JHI=m\angle JGI+m\ angle GJH[/tex] ----> equation A

Remember that

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

so

[tex]m\angle JHI=\frac{b}{2}[/tex]

[tex]m\angle GJH=\frac{a}{2}[/tex]

substitute in the equation A

[tex]\frac{b}{2}=m\angle JGI+\frac{a}{2}[/tex]

Using the subtraction property

[tex]m\angle JGI=\frac{b}{2}-\frac{a}{2}[/tex]

simplify

[tex]m\angle JGI=\frac{1}{2}(b-a)^o[/tex] ----> proved

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