Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Proof:
Let's consider a circle with center O and radius r. Let P be any point on the circumference of the circle. We need to prove that the tangent line at point P is perpendicular to the radius OP.

Construction:
1. Draw the radius OP from the center O to the point P on the circle.
2. At point P, draw a line perpendicular to OP and label it as PQ.
3. Draw the line segment OQ.

Proof:
To prove that line PQ is perpendicular to OP, we will show that angle OPQ is a right angle.

Step 1: Triangles OQP and OPQ are right triangles.
- Triangle OQP: By construction, OP is a radius of the circle, and OQ is perpendicular to PQ. Therefore, triangle OQP is a right triangle.
- Triangle OPQ: By construction, OP is a radius of the circle, and PQ is perpendicular to OP. Therefore, triangle OPQ is a right triangle.

Step 2: OP is a common side for both triangles OQP and OPQ.
- Since OP is a common side for both triangles, we can conclude that triangles OQP and OPQ share the same hypotenuse.

Step 3: Triangles OQP and OPQ are congruent.
- By the hypotenuse-leg congruence criterion, if two right triangles have the same hypotenuse and one leg of each triangle is congruent, then the triangles are congruent.
- In this case, triangles OQP and OPQ have the same hypotenuse OP and the leg OQ is congruent to itself.
- Therefore, triangles OQP and OPQ are congruent.

Step 4: Angle OPQ is congruent to angle OQP.
- Since triangles OQP and OPQ are congruent, their corresponding parts are congruent.
- Angle OPQ and angle OQP are corresponding angles in triangles OQP and OPQ, respectively.
- Therefore, angle OPQ is congruent to angle OQP.

Step 5: Angle OPQ = angle OQP = 90 degrees.
- In a right triangle, the sum of the measures of the two acute angles is 90 degrees.
- Since angle OPQ and angle OQP are congruent and both are acute angles, their measures must be equal.
- Therefore, angle OPQ = angle OQP = 90 degrees.

Conclusion: Since angle OPQ is a right angle, line PQ is perpendicular to line OP. Hence, the tangent at point P is perpendicular to the radius OP.

This completes the proof of theorem 10.1: "The tangent at any point of a circle is perpendicular to the radius through the point of contact."

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