Power of a Point Theorem: Difference between revisions

Latest revision as of 00:33, 21 October 2025

The Power of a Point Theorem is a relationship that holds between the lengths of the line segments formed when two lines intersect a circle and each other.

Statement

There are three unique cases for this theorem. Each case expresses the relationship between the length of line segments that pass through a common point and touch a circle in at least one point. Can be useful with cyclic quadrilaterals as well however with a slightly different application.

Case 1 (Inside the Circle):

If two chords $AB$ and $CD$ intersect at a point $P$ within a circle, then $AP\cdot BP=CP\cdot DP$

$[asy] draw(circle((0,0),3)); dot((-2.82,1)); label("A",(-3.05,1.25)); dot((1,2.828)); label("B",(1.25,3.05)); draw((-2.82,1)---(1,2.828)); dot((2.3,-1.926)); label("C",(2.55,-2.346)); dot((-2.12,2.123)); label("D",(-2.37,2.507)); draw((2.3,-1.926)---(-2.12,2.123)); dot((-1.556,1.602)); label("P",(-1.656,1.202)); [/asy]$

Case 2 (Outside the Circle):

Classic Configuration

Given lines $BP$ and $CP$ originate from two unique points on the circumference of a circle ( $B$ and $C$ ), intersect each other at point $P$ , outside the circle, and re-intersect the circle at points $A$ and $D$ respectively, then $PA\cdot PB=PD\cdot PC$

$[asy] draw(circle((0,0),3)); dot((1.5,2.598)); label("B",(2,3)); label("P",(-6,1.6)); dot((-6,1)); label("C",(2.55,-2.5)); dot((2.12,-2.123)); dot((-2.996,-0.155)); label("D",(-3.350, -0.6)); dot((-2.429,1.761)); label("A",(-2.729,2.061)); draw((1.5,2.598)---(-6,1)); draw((2.12,-2.123)---(-6,1)); [/asy]$

Tangent Line

Given Lines $AB$ and $AC$ with $AC$ tangent to the related circle at $C$ , $A$ lies outside the circle, and Line $AB$ intersects the circle between $A$ and $B$ at $D$ , $AD\cdot AB=AC^{2}$

$[asy] draw(circle((0,0),3)); dot((0,3)); label("C",(0,3.5)); dot((-8,3)); label("A",(-8,3.5)); dot((2.5,-1.658)); label("B",(2.8,-1.958)); draw((0,3)---(-8,3)); draw((2.5,-1.658)---(-8,3)); dot((-2.907,0.741)); label("D",(-3.357,0.421)); [/asy]$

Case 3 (On the Border/Useless Case):

If two chords, $AB$ and $AC$ , have $A$ on the border of the circle, then the same property such that if two lines that intersect and touch a circle, then the product of each of the lines segments is the same. However since the intersection points lies on the border of the circle, one segment of each line is $0$ so no matter what, the constant product is $0$ .

$[asy] draw(circle((0,0),3)); dot((1,2.828)); label("A",(1.4,3.028)); dot((-2.5,-1.658)); label("B",(-2.8,-1.958)); dot((2.04,-2.2)); label("C",(2.34,-2.5)); draw((1,2.828)---(-2.5,-1.658)); draw((1,2.828)---(2.04,-2.2)); [/asy]$

Alternate Formulation

This alternate formulation is much more compact, convenient, and general.

Consider a circle $O$ and a point $P$ in the plane where $P$ is not on the circle. Now draw a line through $P$ that intersects the circle in two places. The power of a point theorem says that the product of the length from $P$ to the first point of intersection and the length from $P$ to the second point of intersection is constant for any choice of a line through $P$ that intersects the circle. This constant is called the power of point $P$ . For example, in the figure below $PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i$

Hint for Proof (Stop here if you want to prove this yourself)

Draw extra lines to create similar triangles (Draw $AD$ on all three figures. Draw another line as well.)

Notice how this definition still works if $A_k$ and $B_k$ coincide (as is the case with $X$ ). Consider also when $P$ is inside the circle. The definition still holds in this case.

Notes

One important result of this theorem is that both tangents from any point $P$ outside of a circle to that circle are equal in length.

The theorem generalizes to higher dimensions, as follows.

Let $P$ be a point, and let $S$ be an $n$ -sphere. Let two arbitrary lines passing through $P$ intersect $S$ at $A_1,B_1;A_2,B_2$ , respectively. Then $PA_1\cdot PB_1=PA_2\cdot PB_2$

Proof. We have already proven the theorem for a $1$ -sphere (a circle), so it only remains to prove the theorem for more dimensions. Consider the plane $p$ containing both of the lines passing through $P$ . The intersection of $P$ and $S$ must be a circle. If we consider the lines and $P$ with respect simply to that circle, then we have reduced our claim to the case of two dimensions, in which we know the theorem holds.

Proof

Case 1 (Inside the Circle)

Join $AD$ and $BC$ .

In $\triangle ADP \; \text{and} \; \triangle CBP$

$\angle ADC = \angle CBA \hspace{1cm}$ (Angles subtended by the same segment are equal)

$\angle DPA = \angle BPC \hspace{1cm}$ (Vertically opposite angles)

$\therefore \; \triangle ADP \sim \triangle CBP$

$\implies \frac{AP}{CP} = \frac{DP}{BP} \hspace{1cm}$ (Corresponding sides of similar triangles are in the same ratio)

$\implies AP \cdot BP = DP \cdot CP$

$\blacksquare$

Case 2 (Outside the Circle)

Join $AD$ and $BC$

$\angle DAB + \angle DCB = 180^{\circ} = \angle PAD + \angle DAB \hspace{1cm}$ (Why?)

$\implies \angle PCB = \angle DCB = \angle PAD$

Now, In $\triangle PAD \; \text{and} \; \triangle PCB$

$\angle PAD = \angle PCB \hspace{1cm}$ (shown above)

$\angle APD = \angle CPB \hspace{1cm}$ (common angle)

$\therefore \; \triangle PAD \sim \triangle PCB$

$\implies \frac{PA}{PC} = \frac{PD}{PB} \hspace{1cm}$ (Corresponding sides of similar triangles are in the same ratio)

$\implies PA \cdot PB = PD \cdot PC$

$\blacksquare$

Case 3 (On the Circle Border)

Length of a point is zero so no proof needed :)

Problems

Introductory

Find the value of $x$ in the following diagram:

Solution

Find the value of $x$ in the following diagram:

Solution

(ARML) In a circle, chords $AB$ and $CD$ intersect at $R$ . If $AR:BR=1:4$ and $CR:DR=4:9$ , find the ratio $AB:CD$ .

Solution

(ARML) Chords $AB$ and $CD$ of a given circle are perpendicular to each other and intersect at a right angle at point $E$ . Given that $BE=16$ , $DE=4$ , and $AD=5$ , find $CE$ .

Solution

Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$ (Source)

Intermediate

Two tangents from an external point $P$ are drawn to a circle and intersect it at $A$ and $B$ . A third tangent meets the circle at $T$ , and the tangents $\overrightarrow{PA}$ and $\overrightarrow{PB}$ at points $Q$ and $R$ , respectively (this means that T is on the minor arc $AB$ ). If $AP = 20$ , find the perimeter of $\triangle PQR$ . (Source)

Square $ABCD$ of side length $10$ has a circle inscribed in it. Let $M$ be the midpoint of $\overline{AB}$ . Find the length of that portion of the segment $\overline{MC}$ that lies outside of the circle. (Source)

$DEB$ is a chord of a circle such that $DE=3$ and $EB=5 .$ Let $O$ be the center of the circle. Join $OE$ and extend $OE$ to cut the circle at $C.$ Given $EC=1,$ find the radius of the circle. (Source)

Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$ (Source)

Let $ABC$ be a triangle inscribed in circle $\omega$ . Let the tangents to $\omega$ at $B$ and $C$ intersect at point $D$ , and let $\overline{AD}$ intersect $\omega$ at $P$ . If $AB=5$ , $BC=9$ , and $AC=10$ , $AP$ can be written as the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime integers. Find $m + n$ . (Source)

Olympiad

Given circles $\omega_1$ and $\omega_2$ intersecting at points $X$ and $Y$ , let $\ell_1$ be a line through the center of $\omega_1$ intersecting $\omega_2$ at points $P$ and $Q$ and let $\ell_2$ be a line through the center of $\omega_2$ intersecting $\omega_1$ at points $R$ and $S$ . Prove that if $P, Q, R$ and $S$ lie on a circle then the center of this circle lies on line $XY$ .

(Source)

Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$ ). The lines $AP$ , $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$ , $L$ , respectively $M$ . The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$ . Show that from $SC = SP$ follows $MK = ML$ .

(Source)

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Power of a Point Theorem: Difference between revisions

Latest revision as of 00:33, 21 October 2025

Contents

Statement

Case 1 (Inside the Circle):

Case 2 (Outside the Circle):

Classic Configuration

Tangent Line

Case 3 (On the Border/Useless Case):

Alternate Formulation

Hint for Proof (Stop here if you want to prove this yourself)

Notes

Proof

Case 1 (Inside the Circle)

Case 2 (Outside the Circle)

Case 3 (On the Circle Border)

Problems

Introductory

Intermediate

Olympiad

See Also

External Links

@@ Line 1: / Line 1: @@
-The '''Power of a Point Theorem''' expresses the relation between the lengths involved with the intersection of two lines between each other and their intersections with a [[circle]].
+The '''Power of a Point Theorem''' is a relationship that holds between the lengths of the [[line segment]]s formed when two [[line]]s [[intersect]] a [[circle]] and each other.
-== Theorem ==
+== Statement ==
-There are three possibilities as displayed in the figures below.
+There are three unique cases for this theorem. Each case expresses the relationship between the length of line segments that pass through a common point and touch a circle in at least one point. Can be useful with [[cyclic quadrilaterals]] as well however with a slightly different application.
-# The two lines are [[secant line|secants]] of the circle and intersect inside the circle (figure on the left).  In this case, we have <math> AE\cdot CE = BE\cdot DE </math>.
+===Case 1 (Inside the Circle):===
-# One of the lines is [[tangent line|tangent]] to the circle while the other is a [[secant line|secant]] (middle figure).  In this case, we have <math> AB^2 = BC\cdot BD </math>.
-# Both lines are [[secant line|secants]] of the circle and intersect outside of it (figure on the right).  In this case, we have <math> CB\cdot CA = CD\cdot CE. </math>
-[[Image:Pop.PNG|center]]
+If two chords <math> AB </math> and <math> CD </math> intersect at a point <math> P </math> within a circle, then <math> AP\cdot BP=CP\cdot DP </math>
+<asy> draw(circle((0,0),3));
+dot((-2.82,1));
+label("A",(-3.05,1.25));
+dot((1,2.828));
+label("B",(1.25,3.05));
+draw((-2.82,1)---(1,2.828));
+dot((2.3,-1.926));
+label("C",(2.55,-2.346));
+dot((-2.12,2.123));
+label("D",(-2.37,2.507));
+draw((2.3,-1.926)---(-2.12,2.123));
+dot((-1.556,1.602));
+label("P",(-1.656,1.202));
+</asy>
+===Case 2 (Outside the Circle):===
+=====Classic Configuration=====
+Given lines <math> BP </math> and <math> CP </math> originate from two unique points on the [[circumference]] of a circle (<math> B </math> and <math> C </math>), intersect each other at point <math> P </math>, outside the circle, and re-intersect the circle at points <math> A </math> and <math> D </math> respectively, then <math> PA\cdot PB=PD\cdot PC </math>
+<asy> draw(circle((0,0),3));
+dot((1.5,2.598));
+label("B",(2,3));
+label("P",(-6,1.6));
+dot((-6,1));
+label("C",(2.55,-2.5));
+dot((2.12,-2.123));
+dot((-2.996,-0.155));
+label("D",(-3.350, -0.6));
+dot((-2.429,1.761));
+label("A",(-2.729,2.061));
+draw((1.5,2.598)---(-6,1));
+draw((2.12,-2.123)---(-6,1));
+</asy>
+=====Tangent Line=====
+Given Lines <math> AB </math> and <math> AC </math> with <math> AC </math> [[tangent line|tangent]] to the related circle at <math> C </math>, <math> A </math> lies outside the circle, and Line <math> AB </math> intersects the circle between <math> A </math> and <math> B </math> at <math> D </math>, <math> AD\cdot AB=AC^{2} </math>
+<asy> draw(circle((0,0),3));
+dot((0,3));
+label("C",(0,3.5));
+dot((-8,3));
+label("A",(-8,3.5));
+dot((2.5,-1.658));
+label("B",(2.8,-1.958));
+draw((0,3)---(-8,3));
+draw((2.5,-1.658)---(-8,3));
+dot((-2.907,0.741));
+label("D",(-3.357,0.421));
+</asy>
+===Case 3 (On the Border/Useless Case):===
+If two chords, <math> AB </math> and <math> AC </math>, have <math> A </math> on the border of the circle, then the same property such that if two lines that intersect and touch a circle, then the product of each of the lines segments is the same. However since the intersection points lies on the border of the circle, one segment of each line is <math> 0 </math> so no matter what, the constant product is <math> 0 </math>.
+<asy> draw(circle((0,0),3));
+dot((1,2.828));
+label("A",(1.4,3.028));
+dot((-2.5,-1.658));
+label("B",(-2.8,-1.958));
+dot((2.04,-2.2));
+label("C",(2.34,-2.5));
+draw((1,2.828)---(-2.5,-1.658));
+draw((1,2.828)---(2.04,-2.2));
+</asy>
 === Alternate Formulation ===
 This alternate formulation is much more compact, convenient, and general.
-Consider a circle O and a point P in the plane where P is not on the circle.  Now draw a line through P that intersects the circle in two places.  The power of a point theorem says that the product of the length from P to the first point of intersection and the length from P to the second point of intersection is constant for any choice of a line through P that intersects the circle.  This constant is called the power of point P.  For example, in the figure below
+Consider a circle <math>O</math> and a point <math>P</math> in the plane where <math>P</math> is not on the circle. Now draw a line through <math>P</math> that intersects the circle in two places. The power of a point theorem says that the product of the length from <math>P</math> to the first point of intersection and the length from <math>P</math> to the second point of intersection is constant for any choice of a line through <math>P</math> that intersects the circle. This constant is called the power of point <math>P</math>. For example, in the figure below
-<cmath>
+<cmath>PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i</cmath>
-PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i
-</cmath>
 [[Image:Popalt.PNG|center]]
-Notice how this definition still works if <math>A_k</math> and <math>B_k</math> coincide (as is the case with X).  Consider also when P is inside the circle.  The definition still holds in this case.
+=== Hint for Proof (Stop here if you want to prove this yourself) ===
+Draw extra lines to create similar triangles (Draw <math>AD</math> on all three figures. Draw another line as well.)
+Notice how this definition still works if <math>A_k</math> and <math>B_k</math> coincide (as is the case with <math>X</math>). Consider also when <math>P</math> is inside the circle. The definition still holds in this case.
+== Notes ==
-== Additional Notes ==
+One important result of this theorem is that both tangents from any point <math>P</math> outside of a circle to that circle are equal in length.
-One important result of this theorem is that both tangents from a point <math> P </math> outside of a circle to that circle are equal in length.
 The theorem generalizes to higher dimensions, as follows.
-Let <math>P</math> be a point, and let <math>S</math> be an <math>n</math>-sphere. Let two arbitrary lines passing through <math>P </math> intersect <math>S </math> at <math>A_1 , B_1 ; A_2 , B_2 </math>, respectively. Then
+Let <math>P</math> be a point, and let <math>S</math> be an <math>n</math>-sphere. Let two arbitrary lines passing through <math>P</math> intersect <math>S</math> at <math>A_1,B_1;A_2,B_2</math>, respectively. Then
-<cmath>
+<cmath>PA_1\cdot PB_1=PA_2\cdot PB_2</cmath>
-PA_1 \cdot PB_1 = PA_2 \cdot PB_2
-</cmath>
+''Proof.'' We have already proven the theorem for a <math>1</math>-sphere (a circle), so it only remains to prove the theorem for more dimensions.  Consider the [[plane]] <math>p</math> containing both of the lines passing through <math>P</math>.  The intersection of <math>P</math> and <math>S</math> must be a circle.  If we consider the lines and <math>P</math> with respect simply to that circle, then we have reduced our claim to the case of two dimensions, in which we know the theorem holds.
+==Proof==
+===Case 1 (Inside the Circle)===
+Join <math>AD</math> and <math>BC</math>.
+In <math>\triangle ADP \; \text{and} \; \triangle CBP</math>
+<math>\angle ADC = \angle CBA \hspace{1cm}</math>  (Angles subtended by the same segment are equal)
+<math>\angle DPA = \angle BPC \hspace{1cm}</math>  (Vertically opposite angles)
+<math>\therefore \; \triangle ADP \sim \triangle CBP</math>
+<math>\implies \frac{AP}{CP} = \frac{DP}{BP} \hspace{1cm}</math> (Corresponding sides of similar triangles are in the same ratio)
+<math>\implies AP \cdot BP = DP \cdot CP</math>
+<math>\blacksquare</math>
+===Case 2 (Outside the Circle)===
+Join <math>AD</math> and <math>BC</math>
+<math>\angle DAB + \angle DCB = 180^{\circ} = \angle PAD + \angle DAB \hspace{1cm}</math> (Why?)
+<math>\implies \angle PCB = \angle DCB = \angle PAD</math>
+Now, In <math>\triangle PAD \; \text{and} \; \triangle PCB</math>
+<math>\angle PAD = \angle PCB \hspace{1cm}</math> (shown above)
+<math>\angle APD = \angle CPB \hspace{1cm}</math> (common angle)
+<math>\therefore \; \triangle PAD \sim \triangle PCB</math>
+<math>\implies \frac{PA}{PC} = \frac{PD}{PB} \hspace{1cm}</math> (Corresponding sides of similar triangles are in the same ratio)
+<math>\implies PA \cdot PB = PD \cdot PC</math>
+<math>\blacksquare</math>
-''Proof.''  We have already proven the theorem for a 1-sphere (i.e., a circle), so it only remains to prove the theorem for more dimensions.  Consider the [[plane]] <math>p </math> containing both of the lines passing through <math>P </math>.  The intersection of <math>P </math> and <math>S </math> must be a circle.  If we consider the lines and <math>P </math> with respect simply to that circle, then we have reduced our claim to the case of two dimensions, in which we know the theorem holds.
+===Case 3 (On the Circle Border)===
+Length of a point is zero so no proof needed :)
 == Problems ==
-The problems are divided into three categories: introductory, intermediate, and olympiad.
 === Introductory ===
-==== Problem 1 ====
-Find the value of <math> x </math> in the following diagram:
-[[Image:popprob1.PNG|center]]
+* Find the value of <math>x</math> in the following diagram: [[Image:popprob1.PNG|center]]
+:[[Power of a Point Theorem/Introductory_Problem_1|Solution]]
+* Find the value of <math>x</math> in the following diagram: [[Image:popprob2.PNG|center]]
+:[[Power of a Point Theorem/Introductory_Problem_2|Solution]]
+* ([[ARML]]) In a circle, chords <math>AB</math> and <math>CD</math> intersect at <math>R</math>. If <math>AR:BR=1:4</math> and <math>CR:DR=4:9</math>, find the ratio <math>AB:CD</math>   .
+[[Image:popprob3.PNG|center]]
+:[[Power of a Point Theorem/Introductory_Problem_3|Solution]]
+* ([[ARML]]) Chords <math>AB</math> and <math>CD</math> of a given circle are [[perpendicular]] to each other and intersect at a right angle at point <math>E</math>. Given that <math>BE=16</math>, <math>DE=4</math>, and <math>AD=5</math>, find <math>CE</math>.
+:[[Power of a Point Theorem/Introductory_Problem_4|Solution]]
+* Let <math>\overline{AB}</math> be a diameter in a circle of radius <math>5\sqrt2.</math> Let <math>\overline{CD}</math> be a chord in the circle that intersects <math>\overline{AB}</math> at a point <math>E</math> such that <math>BE=2\sqrt5</math> and <math>\angle AEC = 45^{\circ}.</math> What is <math>CE^2+DE^2?</math> ([[2020 AMC 12B Problems/Problem 12|Source]])
+=== Intermediate ===
-[[Power of a Point Theorem/Introductory_Problem_1|Solution]]
+* Two tangents from an external point <math>P</math> are drawn to a circle and intersect it at <math>A</math> and <math>B</math>.  A third tangent meets the circle at <math>T</math>, and the tangents <math>\overrightarrow{PA}</math> and <math>\overrightarrow{PB}</math> at points <math>Q</math> and <math>R</math>, respectively (this means that T is on the minor arc <math>AB</math>). If <math>AP = 20</math>, find the perimeter of <math>\triangle PQR</math>. ([[1961_AHSME_Problems/Problem_11|Source]])
-==== Problem 2 ====
+* Square <math>ABCD</math> of side length <math>10</math> has a circle inscribed in it. Let <math>M</math> be the midpoint of <math>\overline{AB}</math>. Find the length of that portion of the segment <math>\overline{MC}</math> that lies outside of the circle. ([[2020 AMC 12B Problems/Problem 10|Source]])
-Find the value of <math> x </math> in the following diagram.
-[[Image:popprob2.PNG|center]]
+* <math>DEB</math> is a chord of a circle such that <math>DE=3</math> and <math>EB=5 .</math> Let <math>O</math> be the center of the circle. Join <math>OE</math> and extend <math>OE</math> to cut the circle at <math>C.</math> Given <math>EC=1,</math> find the radius of the circle. ([[1971_Canadian_MO_Problems/Problem_1|Source]])
+[[Image:CanadianMO_1971-1.jpg]]
-[[Power of a Point Theorem/Introductory_Problem_2|Solution]]
+* Triangle <math>ABC</math> has <math>BC=20.</math> The incircle of the triangle evenly trisects the median <math>AD.</math> If the area of the triangle is <math>m \sqrt{n}</math> where <math>m</math> and <math>n</math> are integers and <math>n</math> is not divisible by the square of a prime, find <math>m+n.</math> ([[2005 AIME I Problems/Problem 15|Source]])
-==== Problem 3 ====
+* Let <math>ABC</math> be a triangle inscribed in circle <math>\omega</math>. Let the tangents to <math>\omega</math> at <math>B</math> and <math>C</math> intersect at point <math>D</math>, and let <math>\overline{AD}</math> intersect <math>\omega</math> at <math>P</math>. If <math>AB=5</math>, <math>BC=9</math>, and <math>AC=10</math>, <math>AP</math> can be written as the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime integers. Find <math>m + n</math>. ([[2024 AIME I Problems/Problem 10|Source]])
-([[ARML]]) In a circle, chords <math> AB </math> and <math> CD </math> intersect at <math> R </math>. If <math>AR:BR = 1:4 </math> and <math> CR:DR = 4:9 </math>, find the ratio <math> AB:CD. </math>
-[[Image:popprob3.PNG|center]]
+=== Olympiad ===
-[[Power of a Point Theorem/Introductory_Problem_3|Solution]]
+* Given circles <math>\omega_1</math> and <math>\omega_2</math> intersecting at points <math>X</math> and <math>Y</math>, let <math>\ell_1</math> be a line through the center of <math>\omega_1</math> intersecting <math>\omega_2</math> at points <math>P</math> and <math>Q</math> and let <math>\ell_2</math> be a line through the center of <math>\omega_2</math> intersecting <math>\omega_1</math> at points <math>R</math> and <math>S</math>. Prove that if <math>P, Q, R</math> and <math>S</math> lie on a circle then the center of this circle lies on line <math>XY</math>.
-==== Problem 4 ====
+([[2009 USAMO Problems/Problem 1|Source]])
-([[ARML]]) Chords <math> AB </math> and <math> CD </math> of a given circle are perpendicular to each other and intersect at a right angle.  Given that <math> BE = 16, DE = 4, </math> and <math> AD = 5 </math>, find <math> CE </math>.
-[[Power of a Point Theorem/Introductory_Problem_4|Solution]]
+Let <math>P</math> be a point interior to triangle <math>ABC</math> (with <math>CA \neq CB</math>). The lines <math>AP</math>, <math>BP</math> and <math>CP</math> meet again its circumcircle <math>\Gamma</math> at <math>K</math>, <math>L</math>, respectively <math>M</math>. The tangent line at <math>C</math> to <math>\Gamma</math> meets the line <math>AB</math> at <math>S</math>. Show that from <math>SC = SP</math> follows <math>MK = ML</math>.
-=== Intermediate ===
+([[2010 IMO Problems/Problem 4|Source]])
-==== Problem 1 ====
-Two tangents from an external point <math> P </math> are drawn to a circle and intersect it at <math> A </math> and <math> B </math>.  A third tangent meets the circle at <math> T </math>, and the tangents <math> \overrightarrow{PA} </math> and <math> \overrightarrow{PB} </math> at points <math> Q </math> and <math> R </math>, respectively.  Find the perimeter of <math> \triangle PQR </math>.
-==== Other Intermediate Example Problems ====
+== See Also ==
-* [[1971_Canadian_MO_Problems/Problem_1 | 1971 Canadian Mathematics Olympiad Problem 1]]
-==See also==
 * [[Geometry]]
 * [[Planar figures]]
+=== External Links ===
+* [{{SERVER}}/community/c6h2513977 Handout on AoPS Forums]
 [[Category:Geometry]]
 [[Category:Theorems]]
+{{stub}}