Given:
BAL is a right triangle at A.
O is the midpoint of [BL].
The circle of diameter [BO] cuts [BA] at I.
Show that I is the midpoint of [AB].
Here is a second way.
The $\displaystyle \angle OIB$ being inscribed in a semi-circle is a right angle.
Therefore, $\displaystyle \overleftrightarrow {AL}||\overleftrightarrow {BI}$ so by the midpoint theorem $\displaystyle I$ is the midpoint of $\displaystyle \overline {AB} $.