In a circle, two chords \(UU'\) and \(VV'\) are drawn perpendicular to a diameter \(ST\). Determine that the segment \(MM'\) joining the midpoints of the chords \(UV\) and \(U'V'\) is perpendicular to \(ST\).
Proof:
Let \(O\) be the centre of the circle.
Since, the chords \(UU'\) and \(VV'\) are the two chords perpendicular to \(ST\).
This implies, the chords \(UU'\) and \(VV'\) are .
Now, consider the chords \(UV\) and \(U'V'\).
By theorem,
Therefore, \(OM⊥\) .
That is, \(∠OMU = \angle OMV =\)\(^°\) ......(1)
Also, \(OM'⊥ \).
Therefore, \(\angle OM'V' = \angle OM'U'\) \(^°\) ......(2)
From (1) and (2), we can say that, the chords \(UV\) and \(U'V'\) are .
Thus, we can say that, \(M\), \(M'\) and \(O\) are .
\(ST\) passes through . [Given]
Therefore, the line \(MM'\) passes through the centre \(O\) and is perpendicular to \(ST\).
Hence, proved.