Question

proving the first corollary to the inscribed angle theorem
given: ∠kjl and ∠kml intercept arc kl
prove: m∠kjl = m∠kml
statements reasons

1. ∠kjl intercepts kl 1. given
2. ∠kml intercepts kl 2. given
3. m∠kjl = 1/2 (mkl) 3.
4. m∠kml = 1/2 (mkl) 4.
5. m∠kjl = m∠kml 5.

angle formed by a tangent and a chord thm.
inscribed angle thm.
second corollary to the inscribed angle thm.
substitution property

Explanation:

Step1: Recall inscribed - angle theorem

The measure of an inscribed angle is half the measure of the intercepted arc. So, since $\angle KJL$ is an inscribed angle intercepting $\overset{\frown}{KL}$, $m\angle KJL=\frac{1}{2}(m\overset{\frown}{KL})$ by the inscribed - angle theorem.

Step2: Apply inscribed - angle theorem again

Since $\angle KML$ is an inscribed angle intercepting $\overset{\frown}{KL}$, $m\angle KML = \frac{1}{2}(m\overset{\frown}{KL})$ by the inscribed - angle theorem.

Step3: Use substitution property

Since $m\angle KJL=\frac{1}{2}(m\overset{\frown}{KL})$ and $m\angle KML=\frac{1}{2}(m\overset{\frown}{KL})$, we can substitute to get $m\angle KJL=m\angle KML$ by the substitution property.

Answer:

1. inscribed angle thm.
2. inscribed angle thm.
3. substitution property