nyuuzyou/wikis huggingface dataset:
Area of Triangle Inscribed in Parabola/Proof 1

== Theorem ==
== Proof ==
The point $\tuple {x = 0, y = 0}$ satisfies $y = x^2$.
Since $x^2 \ge 0$, $0$ is the minimum value for $y$.
Thus, the vertex of $T$ lies at the origin.
The coordinates of $C$ are given as the average those for $A$ and $B$.
Given:
:$A = \tuple {u, u^2}$
:$B = \tuple {v, v^2}$, with $u > v$
By the definition of average: 
:$C = \tuple {\dfrac 1 2 \paren {u + v}, \dfrac 1 4 \paren {u + v}^2 } $
Let $P$ be the midpoint of $AB$, such that $AP = BP$.
Let the coordinates of $P$ be $\tuple {P_x, P_y}$.
By the Midline Theorem:
:$P_x = \dfrac 1 2 \paren {u + v}$
Also by the Midline Theorem:
:$P_y = \dfrac 1 2 \paren {u^2 + v^2}$
Let $h$ be one-half the distance between the $x$-coordinates of $A$ and $B$:
:$h = \dfrac 1 2 \paren {u - v}$
By construction:
:the height of both $\triangle APC$ and $\triangle BPC$ is equal to $h$.
Let $b$ be the length of $PC$, the shared base of $\triangle APC$ and $\triangle BPC$.
 
By Area of Triangle:
:$\map \AA {\triangle APC} = \dfrac 1 2 h b$
Substituting from above:
:$\map \AA {\triangle APC} = \dfrac 1 {16} \paren {u - v}^3$
Since they have the same height and a shared base:
:$\map \AA {\triangle APC} = \map \AA {\triangle BPC}$
But:
:$\map \AA {\triangle ABC} = \map \AA {\triangle APC} + \map \AA {\triangle BPC}$
By addition:
:$\map \AA {\triangle ABC} = h b = \dfrac 1 8 \paren {u - v}^3$
Category:Area of Triangle Inscribed in Parabola
Area of Triangle Inscribed in Parabola/Proof 2

== Theorem ==
== Proof ==
From Two-Point Form of Equation of Straight Line in Plane, the straight line $AB$ can be expressed as:
 
 
The coordinates of $C$ are: 
:$\tuple {\dfrac 1 2 \paren {u + v}, \dfrac 1 4 \paren {u + v}^2 } $
Let $P$ be the midpoint of $AB$, such that $AP = BP$.
We can use Two-Point Form of Equation of Straight Line in Plane above to determine the coordinates of $P$:
 
 
Therefore, the coordinates of $P$ are: 
:$\tuple {\dfrac 1 2 \paren {u + v}, \dfrac 1 2 \paren {u^2 + v^2} } $
:500px
We now connect points $P$ and $C$ to create line segment $PC$
This in turn creates two triangles.
We see from the diagram above:
:$\map \AA {\triangle ABC} = \map \AA {\triangle APC} + \map \AA {\triangle CPB}$
From Area of Triangle in Terms of Two Sides and Angle, we have:
:$\map \AA {\triangle APC} = \dfrac 1 2 \paren {AP} \paren {PC} \map \sin {\angle APC} $
We now notice that:
 
 
Therefore:
:$\map \AA {\triangle APC} = \dfrac 1 2 \paren {PC} \dfrac {\paren {u - v} } 2$
We also notice:
 
 
Since $AP = PB$, $PC = PC$ and $\map \sin {\angle CPB} = \map \sin {\angle APC}$, then by Area of Triangle in Terms of Two Sides and Angle we have:
:$\map \AA {\triangle APC} = \map \AA {\triangle CPB}$
Therefore:
 
 
All that remains is to determine the length of $CP$:
 
 
Substituting from above:
:$\map \AA {\triangle ABC} = \dfrac 1 2 \paren {u - v} \dfrac 1 4 \paren {u - v}^2$
Therefore:
:$\map \AA {\triangle ABC} = \dfrac 1 8 \paren {u - v}^3$
Category:Area of Triangle Inscribed in Parabola
Area of Triangle in Determinant Form

== Theorem ==
Let $A = \tuple {x_1, y_1}, B = \tuple {x_2, y_2}, C = \tuple {x_3, y_3}$ be points in the Cartesian plane.
The area $\AA$ of the triangle whose vertices are at $A$, $B$ and $C$ is given by:
:$\AA = \dfrac 1 2 \size {\paren {\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} } }$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Examples ==
== Sources ==
* : $\S 10$: Formulas from Plane Analytic Geometry: Area of Triangle with Vertices at $\tuple {x_1, y_1}$, $\tuple {x_2, y_2}$, $\tuple {x_3, y_3}$: $10.10$
* : $\S 8$: Formulas from Plane Analytic Geometry: Area of Triangle with Vertices at $\tuple {x_1, y_1}$, $\tuple {x_2, y_2}$, $\tuple {x_3, y_3}$: $8.10.$
Category:Area of Triangle in Determinant Form
Category:Areas of Triangles
Category:Determinants
Area of Triangle in Determinant Form/Examples/Vertices at (-4-i), (1+2i), (4-3i)

== Example of Area of Triangle in Determinant Form ==
Let $T$ be a triangle embedded in the complex plane with vertices at $\paren {-4 - i}, \paren {1 + 2 i}, \paren {4 - 3 i}$.
The area of $T$ is given by:
:$\map \Area T = 17$
== Proof ==
From Area of Triangle in Determinant Form:
$\map \Area T = \dfrac 1 2 \size {\paren {\begin{vmatrix}
-4 & -1 & 1 \\
1 & 2 & 1 \\
4 & -3 & 1 \\
\end{vmatrix} } }$
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $114$
Category:Area of Triangle in Determinant Form
Area of Triangle in Determinant Form/Proof 1

== Theorem ==
== Proof ==
:
Let $A$, $B$ and $C$ be defined as complex numbers in the complex plane.
The vectors from $C$ to $A$ and from $C$ to $B$ are given by:
:$z_1 = \paren {x_1 - x_3} + i \paren {y_1 - y_3}$
:$z_2 = \paren {x_2 - x_3} + i \paren {y_2 - y_3}$
From Area of Triangle in Terms of Side and Altitude, $\AA$ is half that of a parallelogram contained by $z_1$ and $z_2$.
Thus:
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Dot and Cross Product: $42$
Category:Area of Triangle in Determinant Form
Area of Triangle in Determinant Form/Proof 2

== Theorem ==
== Proof ==
:
Let $A$, $B$ and $C$ be as defined..
Let $O$ be the origin of the Cartesian plane in which $\triangle ABC$ is embedded.
Taking into account the signs of the areas of the various triangles involved:
:$\triangle ABC = \triangle OAB + \triangle OBC + \triangle OCA$
as it is seen that $\triangle OBC$ and $\triangle OCA$ are described in clockwise sense.
From proof 2 of Area of Triangle in Determinant Form with Vertex at Origin:
The result follows.
== Sources ==
* : Chapter $\text I$. Coordinates: $8$. Area of any triangle
Category:Area of Triangle in Determinant Form
Area of Triangle in Determinant Form/Proof 3

== Theorem ==
== Proof ==
:
Let $A$, $B$ and $C$ be defined as $\tuple {x_1, y_1}$, $\tuple {x_2, y_2}$ and $\tuple {x_3, y_3}$ respectively.
From the figure, we see that:
== Sources ==
* : $\text {III}$. Analytical Geometry: The Straight Line: The area of the triangle $ABC$
Category:Area of Triangle in Determinant Form
Area of Triangle in Determinant Form with Vertex at Origin

== Example of Area of Triangle in Determinant Form ==
Let $A = \tuple {0, 0}, B = \tuple {b, a}, C = \tuple {x, y}$ be points in the Cartesian plane.
Let $T$ the triangle whose vertices are at $A$, $B$ and $C$.
Then the area $\AA$ of $T$ is:
:$\map \Area T = \dfrac {\size {b y - a x} } 2$
== Proof 1 ==
== Proof 2 ==
== Example ==
== Sources ==
* : $\text {III}$. Analytical Geometry: The Straight Line: The area of the triangle $ABC$
* : $\text {2-3}$ The Linear Diophantine Equation: Exercise $4$
Category:Area of Triangle in Determinant Form
Area of Triangle in Determinant Form with Vertex at Origin/Example

== Examples of Area of Triangle in Determinant Form with Vertex at Origin ==
Let $A = \tuple {0, 0}, B = \tuple {b, a}, C = \tuple {x_0, y_0}$ be points in the Cartesian plane.
Let $\tuple {x_0, y_0}$ be a solution to the linear diophantine equation:
:$a x - b y = 1$
Let $T$ the triangle whose vertices are at $A$, $B$ and $C$.
Then the area $\AA$ of $T$ is:
:$\map \Area T = \dfrac 1 2$
== Proof ==
Let $A = \tuple {0, 0}, B = \tuple {b, a}, C = \tuple {x_0, y_0}$ be points in the Cartesian plane.
From Area of Triangle in Determinant Form with Vertex at Origin:
:$\map \Area T = \dfrac 1 2 \size {b y - a x}$
for the triangle $T$ whose vertices are at $\tuple {0, 0}, \tuple {b, a}, \tuple {x, y}$.
We have that:
$\size {b y_0 - a x_0} = 1$
Hence the result.
== Sources ==
* : $\text {2-3}$ The Linear Diophantine Equation: Exercise $5$
Category:Area of Triangle in Determinant Form
Area of Triangle in Determinant Form with Vertex at Origin/Proof 2

== Example of Area of Triangle in Determinant Form ==
== Proof ==
Let the polar coordinates of $B$ and $C$ be:
Let $\theta$ be the angle between $AB$ and $AC$.
Then we have:
We can define the area of $\triangle ABC$ as being positive or negative according to the sign of $\dfrac 1 2 \paren {b y - a x}$.
However, if we are interested only in the absolute value of $\triangle ABC$, as in this context, we can report:
:$\map \Area {\triangle ABC} = \dfrac {\size {b y - a x} } 2$
== Sources ==
* : Chapter $\text I$. Coordinates: $6$. Area of a triangle
Category:Area of Triangle in Determinant Form
Area of Triangle in Terms of Circumradius

== Theorem ==
Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Then the area $\AA$ of $\triangle ABC$ is given by:
:$\AA = \dfrac {a b c} {4 R}$
where $R$ is the circumradius of $\triangle ABC$.
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
This can also been seen in some sources presented as:
:$R = \dfrac {a b c} {4 \AA}$
== Sources ==
* : $\text V$. Trigonometry: Formulae $(52)$
Category:Area of Triangle in Terms of Circumradius
Category:Areas of Triangles
Area of Triangle in Terms of Circumradius/Proof 1

== Theorem ==
== Proof ==
:
Let $O$ be the circumcenter of $\triangle ABC$.
Let $\AA$ be the area of $\triangle ABC$.
Let a perpendicular be dropped from $C$ to $AB$ at $E$.
Let $h := CE$.
Then:
Let a diameter $CD$ of the circumcircle be passed through $O$.
By definition of circumradius, $CD = 2 R$.
By Thales' Theorem, $\angle CAD$ is a right angle.
By Angles on Equal Arcs are Equal, $\angle ADC = \angle ABC$.
It follows from Sum of Angles of Triangle equals Two Right Angles that $\angle ACD = \angle ECB$.
Thus by Equiangular Triangles are Similar $\triangle DAC$ and $\triangle BEC$ are similar.
So:
== Sources ==
* 
Category:Area of Triangle in Terms of Circumradius
Area of Triangle in Terms of Exradius

== Theorem ==
Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Let $\rho_a$ be the exradius of $\triangle ABC$ the excircle which is tangent to $a$.
Let $s$ be the semiperimeter of $\triangle ABC$.
Then the area $\AA$ of $\triangle ABC$ is given by:
:$\AA = \rho_a \paren {s - a}$
== Proof ==
== Also presented as ==
Some sources present this result as:
:$\rho_a = \dfrac \AA {s - a}$
== Sources ==
* : $\text V$. Trigonometry: Formulae $(55)$
Category:Area of Triangle in Terms of Exradius
Category:Excircles of Triangles
Category:Areas of Triangles
Area of Triangle in Terms of Exradius/Proof

== Theorem ==
== Proof ==
:
Let $C$ be the excircle of $\triangle ABC$ which is tangent to $a$.
By definition:
:$\rho_a$ is the radius of $C$
:$I_a$ is the center of $C$.
Then we have:
== Sources ==
* : $\text V$. Trigonometry: The ex-circles
Category:Area of Triangle in Terms of Exradius
Area of Triangle in Terms of Inradius

== Theorem ==
Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Then the area $\AA$ of $\triangle ABC$ is given by:
:$\AA = r s$
where:
:$r$ is the inradius of $\triangle ABC$
:$s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.
== Proof ==
== Also presented as ==
This can also been seen in some sources presented as:
:$r = \dfrac \AA s$
== Sources ==
* : $\text V$. Trigonometry: Formulae $(53)$
Category:Area of Triangle in Terms of Inradius
Category:Areas of Triangles
Category:Incircles of Triangles
Area of Triangle in Terms of Inradius/Proof

== Theorem ==
== Proof ==
:
Let $I$ be the incenter of $\triangle ABC$.
Let $r$ be the inradius of $\triangle ABC$.
The total area of $\triangle ABC$ is equal to the sum of the areas of the triangle formed by the vertices of $\triangle ABC$ and its incenter:
:$\AA = \map \Area {\triangle AIB} + \map \Area {\triangle BIC} + \map \Area {\triangle CIA}$
Let $AB$, $BC$ and $CA$ be the bases of $\triangle AIB, \triangle BIC, \triangle CIA$ respectively.
The lengths of $AB$, $BC$ and $CA$ respectively are $c, a, b$.
The altitude of each of these triangles is $r$.
Thus from Area of Triangle in Terms of Side and Altitude:
Thus:
:$\AA = r \dfrac {a + b + c} 2$
That is:
:$\AA = r s$
where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.
== Sources ==
* : $\text V$. Trigonometry: The incircle
* 
Category:Area of Triangle in Terms of Inradius
Area of Triangle in Terms of Inradius and Exradii

== Theorem ==
Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Then the area $\AA$ of $\triangle ABC$ is given by:
:$\AA = \sqrt {r \rho_a \rho_b \rho_c}$
where:
:$r$ is the inradius
:$\rho_a$, $\rho_b$ and $\rho_c$ are the exradii of $\triangle ABC$ $a$, $b$ and $c$ respectively.
== Proof ==
== Also see ==
* Heron's Formula, to which this is closely related.
Category:Excircles of Triangles
Category:Areas of Triangles
Area of Triangle in Terms of Side and Altitude

== Theorem ==
The area of a triangle $\triangle ABC$ is given by:
:$\dfrac {c \cdot h_c} 2 = \dfrac {b \cdot h_b} 2 = \dfrac {a \cdot h_a} 2$
where:
:$a, b, c$ are the sides
:$h_a, h_b, h_c$ are the altitudes from $A$, $B$ and $C$ respectively.
=== Corollary ===
== Proof 1 ==
== Dissection Proof ==
== Motivation ==
== Sources ==
* : $\S 4$: Geometric Formulas: Triangle of Altitude $h$ and Base $b$: $4.5$
* 
* 
* 
* : $\S 7$: Geometric Formulas: Triangle of Altitude $h$ and Base $b$: $7.5.$
* 
* : Appendix $1$: Areas and volumes
* 
* : Appendix $1$: Areas and volumes
Category:Area of Triangle in Terms of Side and Altitude
Category:Areas of Triangles
Area of Triangle in Terms of Side and Altitude/Motivation

== Area of Triangle in Terms of Side and Altitude: Motivation ==
This formula is perhaps the best-known and most useful for determining a triangle's area.
It is usually remembered, and quoted, as half base times height.
== Sources ==
* 
* 
* : Appendix $1$: Areas and volumes
* 
* : Appendix $1$: Areas and volumes
Category:Area of Triangle in Terms of Side and Altitude
Area of Triangle in Terms of Side and Altitude/Proof 1

== Theorem ==
== Proof ==
:
Construct a point $D$ so that $\Box ABDC$ is a parallelogram.
From Opposite Sides and Angles of Parallelogram are Equal:
:$\triangle ABC \cong \triangle DCB$
hence their areas are equal.
The Area of Parallelogram is equal to the product of one of its bases and the associated altitude.
Thus 
where $\paren {XYZ}$ is the area of the plane figure $XYZ$.
A similar argument can be used to show that the statement holds for the other sides.
== Motivation ==
Category:Area of Triangle in Terms of Side and Altitude
Area of Triangle in Terms of Two Sides and Angle

== Corollary to Area of Triangle in Terms of Side and Altitude ==
The area of a triangle $ABC$ is given by:
:$\dfrac 1 2 a b \sin C$
where:
:$a, b$ are two of the sides
:$C$ is the angle of the vertex opposite its other side $c$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(44)$
* : $\S 4$: Geometric Formulas: Geometric Formulas: Triangle of Altitude $h$ and Base $b$: $4.5$
* : $\S 7$: Geometric Formulas: Triangle of Altitude $h$ and Base $b$: $7.5.$
* 
* : Appendix $1$: Areas and volumes
* 
* : Appendix $1$: Areas and volumes
Category:Area of Triangle in Terms of Two Sides and Angle
Category:Area of Triangle in Terms of Side and Altitude
Category:Areas of Triangles
Area of Triangle in Terms of Two Sides and Angle/Proof 2

== Theorem ==
== Proof ==
:
By definition of sine:
:$h = b \sin C$
From Area of Triangle in Terms of Side and Altitude:
:$\map \Area {ABC} = \dfrac {a h} 2$
Substituting:
:$\map \Area {ABC} = \dfrac {a b \sin C} 2$
== Sources ==
* : $\text V$. Trigonometry: Area of the triangle
Category:Area of Triangle in Terms of Two Sides and Angle
Area of Zone of Sphere

== Theorem ==
Let $\ZZ$ be a zone of a sphere $\SS$.
The area $\AA$ of $\ZZ$ is given by:
:$\AA = 2 \pi R h$
where:
:$R$ is the radius of $\SS$
:$h$ is the height of $\ZZ$.
== Proof ==
== Sources ==
* : $\S 4$: Geometric Formulas: Spherical Cap of Radius $r$ and Height $h$: $4.41$
* 
* 
* : $\S 7$: Geometric Formulas: Spherical Cap of Radius $r$ and Height $h$: $7.41.$
Category:Zones of Spheres
Category:Area Formulas
Area under Acceleration-Time Graph

== Theorem ==
Let $P$ be a particle moving in a straight line.
Let the acceleration of $P$ as a function of time be $\map a t$.
Let the velocity of $P$ as a function of time be $\map v t$.
Let the motion of $P$ be plotted on an acceleration-time-graph $G$.
Let $v_1$ and $v_2$ be the velocities of $P$ at times $t_1$ and $t_2$ respectively.
Then the area under $G$ between times $t_1$ and $t_2$ is equal to $\map {v_2} t = \map {v_1} t$.
== Proof ==
== Sources ==
* 
* 
Category:Acceleration-Time Graphs
Area under Arc of Cycloid

== Theorem ==
Let $C$ be a cycloid generated by the equations:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
Then the area under one arc of the cycloid is $3 \pi a^2$.
That is, the area under one arc of the cycloid is three times the area of the generating circle.
== Proof ==
Let $A$ be the area under of one arc of the cycloid. 
From Area under Curve, $A$ is defined by:
But:
:$\dfrac {\d x} {\d \theta} = a \paren {1 - \cos \theta}$
and so:
== Also see ==
* Length of Arc of Cycloid
== Historical Note ==
== Sources ==
* : Solutions: $209$. -- A Wheel Fallacy
* : Answers: $293$. A Wheel Fallacy
* : $\S 11$: Special Plane Curves: Cycloid: $11.6$
* : $1$: The Nature of Differential Equations: $\S 6$: The Brachistochrone. Fermat and the Bernoullis
* : Chapter $\text {A}.15$: Torricelli ()
* : Chapter $\text {B}.21$: The Cycloid: Example $1$
* : $\S 9$: Special Plane Curves: Cycloid: $9.6.$
Category:Cycloids
Area under Arc of Cycloid/Historical Note

== Historical Note on Area under Arc of Cycloid ==
The proof of the Area under Arc of Cycloid was published in $1644$ by .
, by cutting out templates and measuring their masses, had previously estimated the area as being approximately $3$ times the area of the generating circle.
It is believed that found this result earlier than , but lost priority when he failed to publish his proof.
== Sources ==
* : Chapter $\text {A}.15$: Torricelli ()
* : Chapter $\text {B}.21$: The Cycloid
* 
* 
Category:Historical Notes
Area under Curve

== Theorem ==
Let $f: \R \to \R$ be a real function which is defined and (Darboux) integrable on the closed interval $\closedint a b$.
Consider the graph of $f$ embedded in a Cartesian plane.
Let $\AA$ denote the area between the curve $\map f x$, the straight lines $x = a$ and $x = b$, and the $x$-axis.
Then:
:$\AA = \ds \int_a^b \map f x \rd x$
where $\ds \int_a^b \map f x \rd x$ denotes the (Darboux) definite integral of $f$ from $a$ to $b$.
== Proof ==
:
Let $x \in \closedint a b$.
Let $\delta x$ be an arbitrarily small positive real number such that $x + \delta x \in \closedint a b$.
Consider the small strip between:
:the $x$-axis
:the vertical straight lines through $\tuple {x, 0}$ and $\tuple {x + \delta x, 0}$
:the curve $\map f x$.
The area of this strip is approximated by $y \delta x$.
Hence we create a model of the geometric interpretation of the definite integral.
== Examples ==
== Sources ==
* : $\text {II}$. Calculus: Area, Volume and Centre of Gravity
Category:Area under Curve
Category:Definite Integrals
Category:Analytic Geometry
Area under Curve/Examples

== Examples of Use of Area under Curve ==
=== Area under $y = \paren {x - 1} \paren {x - 2}$ ===
=== Area under $y = \sin x$ between $0$ and $\pi$ ===
=== Area between $x = y^2$ and $x^2 = 8 y$ ===
Category:Examples of Use of Area under Curve
Area under Curve/Examples/(x - 1) (x - 2)

== Example of Use of Area under Curve ==
The area between the $x$-axis and the curve $y = \paren {x - 1} \paren {x - 2}$ is $\dfrac 1 6$.
== Proof ==
Let $\AA$ be the area in question.
The curve $y = \paren {x - 1} \paren {x - 2}$ intercepts the $x$-axis where $y = 0$.
That is, where $x - 1 = 0$ and $x - 2 = 0$, which is $\tuple {1, 0}$ and $\tuple {2, 0}$.
:
Thus from Area under Curve we need to evaluate the definite integral:
:$\AA = \ds \int_1^2 \paren {x - 1} \paren {x - 2} \rd x$
Between those limits, $y$ is negative.
Hence we expect a negative result, which must then be made positive.
So:
== Sources ==
* : $\text {II}$. Calculus: Exercises $\text {XV}$: $2$
Category:Examples of Use of Area under Curve
Area under Curve/Examples/sin x from 0 to pi

== Example of Use of Area under Curve ==
The area bounded by the curve $y = \sin x$ and the $x$-axis between $x = 0$ and $x = \pi$ is $2$.
== Proof ==
Let $\AA$ be the area in question.
:
From Area under Curve we need to evaluate the definite integral:
:$\AA = \ds \int_0^\pi \sin x \rd x$
So:
== Sources ==
* : $\text {II}$. Calculus: Exercises $\text {XV}$: $3$
Category:Examples of Use of Area under Curve
Areas of Circles are as Squares on Diameters

== Theorem ==
=== Lemma ===
== Proof ==
:
Let $ABCD$ and $EFGH$ be circles.
Let $BD$ be a diameter of $ABCD$.
Let $FH$ be a diameter of $EFGH$.
It is to be demonstrated that the ratio of the area of $ABCD$ to the area of $EFGH$ equals the ratio of the square on $BD$ to the square on $FH$.
Suppose to the contrary that the square on $BD$ to the square on $FH$ does not equal the area of $ABCD$ to the area of $EFGH$.
Then:
:$BD^2 : FH^2 = ABCD : S$
where $S \ne EFGH$.
Suppose WLOG that $S < EFGH$.
Let the square $EFGH$ be inscribed in the circle $EFGH$.
From Square Inscribed in Circle is greater than Half Area of Circle:
:the square $EFGH$ is greater than half the circle $EFGH$.
Let the arcs $EF, FG, GH, HE$ be bisected at $K, L, M, N$.
Let $EK, KF, FL, LG, GM, MH, HN, NE$ be joined.
Let the parallelogram be completed on $\triangle EKF$.
Then $\triangle EKF$ is half of this parallelogram.
Therefore the segment $EKF$ of the circle $EFGH$ is less than twice $\triangle EKF$.
Hence each of $\triangle EKF, \triangle FLG, \triangle GMH, \triangle HNE$ is greater than half the segment of the circle $EFGH$ around it.
This process can be repeated.
We continue to bisecting the remaining arcs and joining straight lines repeatedly.
By :
: eventually we will leave some segments of the circle $EFGH$ which will be less than the amount by which the circle $EFGH$ exceeds $S$.
Therefore the remainder, the polygon $EKFLGMHN$ is greater than the area $S$.
Let the polygon $AOBPCQDR$ similar to $EKFLGMHN$ be inscribed in the circle $ABCD$.
Therefore from :
:$BD^2 : FH^2 = AOBPCQDR : EKFLGMHN$
But:
:$BD^2 : FH^2 = ABCD : S$
So by :
:$ABCD : S = AOBPCQDR : EKFLGMHN$
So by :
:$ABCD : AOBPCQDR = S : EKFLGMHN$
But:
:$ABCD > AOBPCQDR$
Therefore:
:$S > EKFLGMHN$
But we started with the premise that:
:$S < EKFLGMHN$
Therefore the square on $BD$ to the square on $FH$ does not equal the area of $ABCD$ to any area less than $EFGH$.
Using the same technique we can also show that the square on $FH$ to the square on $BD$ does not equal the area of $EFGH$ to any area less than $ABCD$.
It is now to be shown that the ratio of the area of $ABCD$ to an area greater than $EFGH$ can equal the ratio of the square on $BD$ to the square on $FH$.
Suppose to the contrary, that:
:$ABCD : S = BD^2 : FH^2$
where $S > EFGH$.
that is:
:$FH^2 : DB^2 = S : ABCD$
But as $S$ is to the circle $ABCD$, so $EFGH$ is to an area less than the circle $ABCD$. 
Therefore by :
:the ratio of the square on $BD$ to the square on $FH$ equals the ratio of the circle $EFGH$ to an area less than the circle $ABCD$.
This was shown to be impossible.
Therefore the square on $BD$ to the square on $FH$ does not equal the area of $ABCD$ to any area greater than $EFGH$.
Hence the result.
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Circles
Areas of Circles are as Squares on Diameters/Lemma

== Theorem ==
== Proof ==
:
Let it be contrived that:
:$S : ABCD = EFGH : T$
for some area $T$.
It is to be demonstrated that $T$ is less than the circle $ABCD$.
From the relationship:
:$S : ABCD = EFGH : T$
it follows from that:
:$S : EFGH = ABCD : T$
But:
:$S > EFGH$
Therefore:
:$ABCD > T$
Hence the result.
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Circles