Step-by-step explanation:
(1)
AX is bisector of [tex] \angle RAT \implies \angle MAN ... (1)[/tex]
[tex] In\: \triangle MIA \: \&\:\triangle NIA\\
\angle AMI \cong \angle ANI ... (each \: 90\degree)\\
\angle MAI \cong \angle NAI ....(from \: 1)\\
AI \cong AI .. (common \:side) \\
\therefore \triangle MIA \: \cong\:\triangle NIA.. \\(By \: AAS \: Postulate \: of \: congruence) \\
\therefore IM = IN[/tex]
(2)
[tex] In \: \square AMIN \\
\angle A = \angle M= \angle N = 90°\\
\therefore \angle I = 90°[/tex]
(Remaining angle of quadrilateral)
[tex] \therefore \square AMIN [/tex] is a rectangle.
[tex] \because IM = IN [/tex]
(adjacent sides of a rectangle)
[tex] \therefore \square AMIN [/tex] is a square.
Hence proved.
