Question

Given: Circle C; Angle LRC is congruent to Angle EIC

Prove: Triangle LRC is congruent to Triangle EIC

Answer 1

`{\triangle LRC} \cong {\triangle EIC}` by angle-angle-side (
`\texttt{AAS}`.)

Explanation:

Let
`r` denote the radius of this circle.

Notice that
`EC` and
`LC` are each a radius of this circle. (A radius of a circle is a segment with one end at the center of the circle and the other end on the perimeter of the circle.)

Therefore, the length of both segment
`EC` and segment
`LC` should be equal to the radius of this circle:

`r = \overline{EC}`.

`r = \overline{LC}`.

Therefore,
`\overline{EC} = \overline{LC}`.

Thus,
`{\triangle LRC} \cong {\triangle EIC}` by angle-angle-side (
`\texttt{AAS}`) since:

• 
  `{\angle RCL} = {\angle ICE}` (same angle.)
• 
  `{\angle LRC} \cong {\angle EIC}` (given.)
• 
  `\overline{EC} = \overline{LC}`.