AB is a chord. Of a circle and the tangents at A, B meet at C. If P is any point on the circle and PL, PM, PN are the perpendiculars from P to AB, BC, CA. Prove that PL^2= PM.PN

To prove that \( PL^2 = PM \cdot PN \), we can use some properties of circles and tangents. This relationship is known as the Power of a Point theorem, which states that for any point outside a circle, the product of the lengths of the segments from that point to the points where tangents touch the circle is equal to the power of the point with respect to the circle. Let's break this down step by step.

Understanding the Setup

We have a circle with a chord \( AB \). The tangents at points \( A \) and \( B \) intersect at point \( C \). Point \( P \) lies on the circumference of the circle, and we drop perpendiculars from \( P \) to the lines \( AB \), \( BC \), and \( CA \), which meet these lines at points \( L \), \( M \), and \( N \) respectively.

Visualizing the Geometry

Imagine the circle with \( AB \) as a horizontal line segment. The points \( A \) and \( B \) are where the tangents touch the circle. The point \( C \) is outside the circle where the two tangents meet. The perpendiculars from point \( P \) create right triangles with the segments \( PL \), \( PM \), and \( PN \) as their heights.

Applying the Power of a Point Theorem

According to the Power of a Point theorem, the power of point \( P \) with respect to the circle can be expressed in terms of the distances from \( P \) to the points where the tangents touch the circle. The theorem states that:

• The square of the length of the tangent from point \( P \) to the circle (let's denote this length as \( PT \)) is equal to the product of the distances from point \( P \) to the points where the tangents intersect the circle.

In our case, the length of the tangent from \( P \) to the circle at points \( A \) and \( B \) can be represented as \( PA \) and \( PB \). Therefore, we can write:

Using the properties of similar triangles formed by the perpendiculars, we can establish that:

Establishing the Relationship

From the right triangles \( PML \) and \( PNL \), we can express the lengths as follows:

• In triangle \( PML \), we have \( PL = PM \cdot \sin(\angle PML) \).
• In triangle \( PNL \), we have \( PL = PN \cdot \sin(\angle PNL) \).

Since \( \angle PML \) and \( \angle PNL \) are complementary (they add up to 90 degrees), we can use the sine relationship to establish that:

Thus, we can equate the two expressions for \( PL \) and rearrange them to show that:

Final Steps to the Proof

Now, squaring both sides of the equation gives us:

Substituting back, we find:

Therefore, we conclude that:

PL² = PM · PN

This relationship holds true for any point \( P \) on the circle, confirming the theorem's validity in this context. The geometric properties of circles, tangents, and perpendiculars work together beautifully to illustrate this fundamental concept in geometry.