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Difference between revisions of "Power of a Point Theorem"

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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.
 
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|chords]] 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>
−
=== Hint for Proof===
+
 
−
Draw extra lines to create similar triangles! (Hint: Draw <math>AD</math> on all three figures. Draw another line as well.)
+
<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 ===
 
=== Alternate Formulation ===
 +
 
This alternate formulation is much more compact, convenient, and general.
 
This alternate formulation is much more compact, convenient, and general.
  
 
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  
 
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]]
 
[[Image:Popalt.PNG|center]]
 +
 +
=== Hint for Proof===
 +
 +
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.
 
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.
  
−
== Additional Notes ==
+
== Notes ==
−
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.
+
 
 +
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.
  
 
The theorem generalizes to higher dimensions, as follows.
 
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)
  
−
''Proof.''  We have already proven the theorem for a <math>1</math>-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.
+
<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>
 +
 
 +
===Case 3 (On the Circle Border)===
 +
 
 +
Length of a point is zero so no proof needed :)
  
 
== Problems ==
 
== Problems ==
−
The problems are divided into three categories: introductory, intermediate, and olympiad.
 
  
 
=== Introductory ===
 
=== 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]]
  
−
[[Power of a Point Theorem/Introductory_Problem_1|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 ===
  
−
==== Problem 2 ==== 
+
* 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]])
−
Find the value of <math>x</math> in the following diagram:
 
  
−
[[Image:popprob2.PNG|center]]
+
* 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]])
  
−
[[Power of a Point Theorem/Introductory_Problem_2|Solution]]
+
* <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]]
  
−
==== Problem 3 ====
+
* 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]])
−
([[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]]
+
* 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]])
  
−
[[Power of a Point Theorem/Introductory_Problem_3|Solution]]
+
=== Olympiad ===
  
−
==== Problem 4 ====
+
* 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>.
−
([[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]]
+
([[2009 USAMO Problems/Problem 1|Source]])
  
−
=== Intermediate ===
+
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>.
−
==== 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 (this means that T is on the minor arc <math>AB</math>). If <math>AB = 20</math>, find the perimeter of <math>\triangle PQR</math>.
 
  
−
==== Problem 2 ====
+
([[2010 IMO Problems/Problem 4|Source]])
−
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.
 
  
−
==== Other Intermediate Example Problems ====
+
== See Also ==
−
* [[1971_Canadian_MO_Problems/Problem_1 | 1971 Canadian Mathematics Olympiad Problem 1]]
 
−
* [[2005 AIME I Problems/Problem 15|2005 AIME I Problem 15]]
 
  
−
==See also==
 
 
* [[Geometry]]
 
* [[Geometry]]
 
* [[Planar figures]]
 
* [[Planar figures]]
 +
 +
=== External Links ===
 +
* [{{SERVER}}/community/c6h2513977 Handout on AoPS Forums]
  
 
[[Category:Geometry]]
 
[[Category:Geometry]]
 
[[Category:Theorems]]
 
[[Category:Theorems]]
 +
{{stub}}

Latest revision as of 22:56, 22 March 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 \[PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i\]

Popalt.PNG

Hint for Proof

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:
    Popprob1.PNG
Solution
  • Find the value of $x$ in the following diagram:
    Popprob2.PNG
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 $AB:CD$ .
Popprob3.PNG
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)

CanadianMO 1971-1.jpg

  • 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)

See Also

External Links

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