Answer:
[tex]{\triangle LRC} \cong {\triangle EIC}[/tex] by angle-angle-side ([tex]\texttt{AAS}[/tex].)
Step-by-step explanation:
Let [tex]r[/tex] denote the radius of this circle.
Notice that [tex]EC[/tex] and [tex]LC[/tex] 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 [tex]EC[/tex] and segment [tex]LC[/tex] should be equal to the radius of this circle:
[tex]r = \overline{EC}[/tex].
[tex]r = \overline{LC}[/tex].
Therefore, [tex]\overline{EC} = \overline{LC}[/tex].
Thus, [tex]{\triangle LRC} \cong {\triangle EIC}[/tex] by angle-angle-side ([tex]\texttt{AAS}[/tex]) since:
- [tex]{\angle RCL} = {\angle ICE}[/tex] (same angle.)
- [tex]{\angle LRC} \cong {\angle EIC}[/tex] (given.)
- [tex]\overline{EC} = \overline{LC}[/tex].
