Answer:
In the situation where segment [tex]\overline {AM}[/tex] is perpendicular to segment [tex]\overline {CI}[/tex], such that [tex]\overline {AM}[/tex] is the perpendicular bisector to segment [tex]\overline {CI}[/tex] we have the following two column proof;
Statement      [tex]{}[/tex]          Reason
M is the midpoint of [tex]\overline {CI}[/tex] Â Â [tex]{}[/tex] Â Â Given
[tex]\overline {CM}[/tex] ≅ [tex]\overline {IM}[/tex]     [tex]{}[/tex]            Definition of midpoint
[tex]\overline {AM}[/tex] ≅ [tex]\overline {AM}[/tex]    [tex]{}[/tex]            Reflexive property
∠CMA = ∠IMA = 90°     [tex]{}[/tex]    Definition of perpendicular bisector
ΔCAM ≅ ΔIAM     [tex]{}[/tex]        By the Side-Angle-Side (SAS) rule of congruency
However, where the segment [tex]\overline {AM}[/tex] is not be assumed as perpendicular to segment [tex]\overline {CI}[/tex], then CNBD
Step-by-step explanation: