To prove that ST = UT, we can use the properties of a regular pentagon and the properties of tangents to a circle.
Given:
1. PQRST is a regular pentagon.
2. QR and PT are tangents to the circle with center O.
3. RSU is a straight line.
Since PQRST is a regular pentagon, all its sides are congruent. Therefore, PQ = QR = RS = ST = TP.
Since QR and PT are tangents to the circle, by the tangent-secant theorem, the length of the tangent segments from an external point to a circle are equal. Therefore, QS = TP.
Now, consider triangle QST.
QS = TP (tangent segments)
QT = QT (common side)
∠QST = ∠QST (common angle)
By the Side-Angle-Side (SAS) congruence criterion, triangle QST is congruent to triangle TQP.
Therefore, ST = UT.
So, we have proved that ST = UT.
