## To prove that PT is perpendicular to OT using the corollary to the Intersecting Secants Property, we need to show that PT^2 = PA * PB.

Looking at the given diagram, we have AB as the diameter of the circle with center O. P is on the extended line of BA, and PT is tangent to the circle at point T.

According to the corollary, we need to compare the product of the lengths of PA and PB (PA * PB) with the square of the length of PT (PT^2).

First, let's analyze the triangle POT:

Since AB is a diameter of the circle, angle OAB is a right angle (90 degrees).

By the definition of a tangent line, angle PTB is also a right angle (90 degrees).

Therefore, we have two right triangles, POT and PTB.

Now, let's examine the triangles PAB and POT:

Both triangles share a common side, PT, and they have two pairs of congruent angles:

- angle PAB is congruent to angle POT (both are right angles)

- angle PBA is congruent to angle PTB (both are shared angles formed by the secant and tangent lines with respect to AB)

By the Angle-Angle Similarity Postulate, we can conclude that triangle PAB and triangle POT are similar.

Because the two triangles are similar, we can establish the following ratio equalities:

PA/PT = PT/PO (from the similar triangles PAB and POT)

PB/PT = PT/PO (from the similar triangles PAB and POT)

Multiplying both ratios by PT, we get:

PA = (PT^2)/PO

PB = (PT^2)/PO

Now, let's substitute these values back into the corollary:

PA * PB = ((PT^2)/PO) * ((PT^2)/PO)

= (PT^2 * PT^2)/(PO * PO)

= (PT^4)/(PO^2)

Comparing this with the expression for PT^2, we have:

PT^2 = (PT^2)/(PO^2)

Therefore, PT^2 = PA * PB, which satisfies the conditions of the corollary to the Intersecting Secants Property.

Hence, we have proved that PT is perpendicular to OT using the corollary to the Intersecting Secants Property.