Question Statement

From a point $P$ lying outside the circle, a tangent $AP$ and a secant line $PBD$ are drawn. Prove that: $|\overline{AP}{|}^2=|\overline{PB}|\cdot |\overline{PD}|$

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Background and Explanation

This proof relies on the properties of right-angled triangles and circle geometry. Specifically, we use the Pythagorean theorem and the property that a perpendicular line drawn from the center of a circle to a chord bisects that chord.

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Solution

To prove the theorem, we first perform a construction. Let $C$ be the center of the circle. Draw a perpendicular line from $C$ to the chord $\overline{BD}$, meeting it at point $E$. By circle properties, $E$ is the midpoint of $B$ and $D$, meaning: $|\overline{BE}|=|\overline{ED}|$

Next, join point $P$ to $C$, and point $B$ to $C$. Let $r$ be the radius of the circle ($|\overline{CA}|=|\overline{CB}|=r$).

Step 1: Apply Pythagorean Theorem to $△APC$

Since $\overline{AP}$ is a tangent, it is perpendicular to the radius $\overline{CA}$. In the right-angled triangle $APC$: $|\overline{CP}{|}^2=|\overline{AP}{|}^2+|\overline{CA}{|}^2$

Step 2: Apply Pythagorean Theorem to $△CPE$

In the right-angled triangle $CPE$ (where $\angle E=90^{\circ }$): $|\overline{CP}{|}^2=|\overline{CE}{|}^2+|\overline{PE}{|}^2$

Step 3: Equate the expressions for $|\overline{CP}{|}^2$

From equations (i) and (ii), we can set the right-hand sides equal to each other: $|\overline{AP}{|}^2+|\overline{CA}{|}^2=|\overline{CE}{|}^2+|\overline{PE}{|}^2$

Step 4: Substitute for $|\overline{CE}{|}^2$ using $△CBE$

In the right-angled triangle $CBE$: $\begin{array}{rl}|\overline{CE}{|}^2+|\overline{BE}{|}^2& =|\overline{CB}{|}^2\\ |\overline{CE}{|}^2& =|\overline{CB}{|}^2-|\overline{BE}{|}^2\end{array}$ Substitute this expression for $|\overline{CE}{|}^2$ into equation (iii): $|\overline{AP}{|}^2+|\overline{CA}{|}^2=(|\overline{CB}{|}^2-|\overline{BE}{|}^2)+|\overline{PE}{|}^2$

Step 5: Simplify using the Radius

Since $|\overline{CA}|$ and $|\overline{CB}|$ are both radii of the circle ($r$), their squares are equal ($|\overline{CA}{|}^2=|\overline{CB}{|}^2$). Subtracting this from both sides: $|\overline{AP}{|}^2=-|\overline{BE}{|}^2+|\overline{PE}{|}^2$ $|\overline{AP}{|}^2=|\overline{PE}{|}^2-|\overline{BE}{|}^2$

Step 6: Expand and Factorize

Note that $|\overline{PE}|=|\overline{PB}|+|\overline{BE}|$. Substituting this: $\begin{array}{rl}|\overline{AP}{|}^2& =(|\overline{PB}|+|\overline{BE}|{)}^2-|\overline{BE}{|}^2\\ & =(|\overline{PB}{|}^2+|\overline{BE}{|}^2+2|\overline{PB}|\cdot |\overline{BE}|)-|\overline{BE}{|}^2\\ & =|\overline{PB}{|}^2+2|\overline{PB}|\cdot |\overline{BE}|\\ & =|\overline{PB}|(|\overline{PB}|+2|\overline{BE}|)\end{array}$

Since $E$ is the midpoint of $BD$, $2|\overline{BE}|=|\overline{BD}|$. Therefore: $\begin{array}{rl}|\overline{AP}{|}^2& =|\overline{PB}|(|\overline{PB}|+|\overline{BD}|)\\ & =|\overline{PB}|\cdot |\overline{PD}|\end{array}$ This completes the proof.

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Key Formulas or Methods Used

• Pythagorean Theorem: $a^2+b^2=c^2$ for right-angled triangles.
• Tangent-Radius Property: A tangent at any point of a circle is perpendicular to the radius through the point of contact.
• Chord Property: A perpendicular from the center of a circle to a chord bisects the chord.
• Segment Addition: $|\overline{PD}|=|\overline{PB}|+|\overline{BD}|$.

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Summary of Steps

1. Construct a perpendicular from center $C$ to chord $BD$ at point $E$.
2. Express the squared hypotenuse $|\overline{CP}{|}^2$ using two different right triangles ($△APC$ and $△CPE$).
3. Equate the two expressions and substitute the radius squared to simplify the equation.
4. Use the Pythagorean theorem on $△CBE$ to eliminate the internal segment $|\overline{CE}{|}^2$.
5. Expand the algebraic expression for the segments along the secant line.
6. Factorize the result to show it equals the product of the external part and the whole secant.