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Question
use the information given in the diagram to prove that (mangle jgi=\frac{1}{2}(b - a)), where (a) and (b) represent the degree measures of arcs fh and ji. angles jhi and gjh are inscribed angles. we have that (mangle jhi=\frac{1}{2}b) and (mangle gjh=\frac{1}{2}a) by the inscribed - angle theorem. 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, (mangle jhi = mangle jgi+mangle gjh). by the, (\frac{1}{2}b=mangle jgi+\frac{1}{2}a). using the subtraction property, (mangle jgi=\frac{1}{2}b-\frac{1}{2}a). therefore, (mangle jgi=\frac{1}{2}(b - a)) by the distributive property.
Step1: Apply inscribed - angle theorem
By the inscribed - angle theorem, for an inscribed angle in a circle, the measure of an inscribed angle is half the measure of its intercepted arc. So, $m\angle{JHI}=\frac{1}{2}b$ (intercepts arc $JI$) and $m\angle{GJH}=\frac{1}{2}a$ (intercepts arc $FH$).
Step2: Use exterior - angle property
$\angle{JHI}$ is an exterior angle of $\triangle{GJH}$. According to the exterior - angle property of a triangle (the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles), we have $m\angle{JHI}=m\angle{JGI}+m\angle{GJH}$.
Step3: Substitute angle measures
Substitute $m\angle{JHI}=\frac{1}{2}b$ and $m\angle{GJH}=\frac{1}{2}a$ into the exterior - angle equation: $\frac{1}{2}b = m\angle{JGI}+\frac{1}{2}a$.
Step4: Solve for $m\angle{JGI}$
Using the subtraction property of equality, subtract $\frac{1}{2}a$ from both sides of the equation: $m\angle{JGI}=\frac{1}{2}b-\frac{1}{2}a$.
Step5: Apply distributive property
Factor out $\frac{1}{2}$ using the distributive property: $m\angle{JGI}=\frac{1}{2}(b - a)$.
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The proof is completed as shown above, demonstrating that $m\angle{JGI}=\frac{1}{2}(b - a)$ using the inscribed - angle theorem, exterior - angle property of a triangle, subtraction property of equality, and distributive property.