nyuuzyou/wikis huggingface dataset:
Alternating Harmonic Series is Conditionally Convergent

== Theorem ==
The alternating harmonic series:
:$\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^\paren {n - 1} } n = 1 - \frac 1 2 + \frac 1 3 - \frac 1 4 + \cdots$
is conditionally convergent.
== Proof ==
Note first that:
:$\ds \sum_{n \mathop = 1}^\infty \size {\frac {\paren {-1}^\paren {n - 1} } n} = \sum_{n \mathop = 1}^\infty \frac 1 n$
which is divergent by Harmonic Series is Divergent.
Hence by definition $\ds \sum_{n \mathop = 1}^\infty \size {\frac {\paren {-1}^\paren {n - 1} } n}$ is not absolutely convergent.
Next note that $\size {\dfrac 1 n}$ is a basic null sequence:
:$\ds \lim_{n \mathop \to \infty} \dfrac 1 n = 0$
and that:
:$\forall n \in \N_{>0}: \dfrac 1 n > 0$
From Reciprocal Sequence is Strictly Decreasing:
:$\dfrac 1 n > \dfrac 1 {n + 1}$
The result follows from the Alternating Series Test.
== Also see ==
* Newton-Mercator Series for ln 2
== Sources ==
* : $\S 1.2.2$: Summary of convergence tests: Example $4$
* 
* 
* 
* 
* 
* 
Category:Alternating Series
Category:Series
Category:Newton-Mercator Series
Alternating Series Test

== Theorem ==
Let $\sequence {a_n}_{N \mathop \ge 0}$ be a decreasing sequence of positive terms in $\R$ which converges with a limit of zero.
That is, let $\forall n \in \N: a_n \ge 0, a_{n + 1} \le a_n, a_n \to 0$ as $n \to \infty$
Then the series:
:$\ds \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} a_n = a_1 - a_2 + a_3 - a_4 + \dotsb$ 
converges.
== Proof ==
First we show that for each $n > m$, we have $0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb \pm a_n \le a_{m + 1}$.
=== Lemma ===
Therefore for each $n > m$, we have:
:$0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb \pm a_n \le a_{m + 1}$
Now, let $\sequence {s_n}$ be the sequence of partial sums of the series:
:$\ds \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} a_n$
Let $\epsilon > 0$.
Since $a_n \to 0$ as $n \to \infty$:
:$\exists N: \forall n > N: a_n < \epsilon$
But $\forall n > m > N$, we have:
Thus we have shown that $\sequence {s_n}$ is a Cauchy sequence.
The result follows from Convergent Sequence is Cauchy Sequence.
== Also known as ==
== Historical Note ==
== Sources ==
* : $\S 6.13$
* 
* : $\S 1.2.2$: Summary of convergence tests: Theorem $1.5$
* 
* 
* 
Category:Alternating Series Test
Category:Alternating Series
Category:Convergence Tests
Category:Named Theorems
Alternating Series Test/Also known as

== Alternating Series Test: Also known as ==
The Alternating Series Test is also seen referred to as Leibniz's Alternating Series Test, for .
Some sources hyphenate: Alternating-Series Test
== Sources ==
* 
* 
* : $\S 1.2.2$: Summary of convergence tests: Theorem $1.5$
Category:Alternating Series Test
Alternating Series Test/Lemma

== Lemma ==
Let $\sequence {a_n}_{N \mathop \ge 0}$ be a decreasing sequence of positive terms in $\R$.
That is, let $\forall n \in \N: a_n \ge 0, a_{n + 1} \le a_n$.
Then fFor all natural numbers $n, m$ with $n > m$ we have: 
:$\ds \sum_{k \mathop = m + 1}^n \paren {-1}^k a_k \le a_{m + 1}$
== Proof ==
We will show this using the Second Principle of Mathematical Induction.
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb \pm a_n \le a_{m + 1}$
=== Basis for the Induction ===
$\map P 1$ holds, as $a_{m + 1} \ge 0$ by definition and $a_{m + 1} \le a_{m + 1}$.
$\map P 2$ also holds, as $a_{m + 2} \le a_{m + 1}$ and so:
:$0 \le a_{m + 1} - a_{m + 2} \le a_{m + 1}$
This is the basis for the induction.
=== Induction Hypothesis ===
In order to simplify the algebra, let:
:$b_k := a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb \pm a_k$
The $\pm$ signifies the fact that $a_k$ is positive for $k$ odd and negative for $k$ even.
Suppose that $\forall j \le k: \map P j$ holds:
:$0 \le b_j \le a_{m + 1}$
This is the induction hypothesis.
We now show that $\forall k: \map P k \implies \map P {k + 1}$.
=== Induction Step ===
This is the induction step:
Suppose $k$ is odd.
By the induction hypothesis:
:$0 \le b_k \le a_{m + 1}$
Because $\map P k$ holds:
:$0 \le b_{k - 1} + a_k \le a_{m + 1}$
But as $a_k \ge a_{k + 1}$:
:$a_k - a_{k + 1} \ge 0$
and so:
:$0 \le b_{k - 1} + \paren {a_k - a_{k + 1} } = b_{k + 1}$
But as $b_k \le a_{m + 1}$:
:$b_k - a_{k + 1} = b_{k + 1} \le a_{m + 1}$
So:
:$0 \le b_{k + 1} \le a_{m + 1}$
or:
:$0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb - a_{k + 1} \le a_{m + 1}$
Thus for odd $k$ it follows that $\map P k \implies \map P {k + 1}$.
Now suppose $k$ is even.
By the induction hypothesis:
:$0 \le b_k \le a_{m + 1}$
Then:
:$0 \le b_k + a_{k + 1} = b_{k + 1}$
Because $\map P k$ holds:
:$0 \le b_{k - 1} - a_k \le a_{m + 1}$
But as $a_k \ge a_{k + 1}$:
:$a_k - a_{k + 1} \ge 0$
and so:
:$b_{k + 1} = b_{k - 1} - a_k + a_{k + 1} = b_{k - 1} - \paren {a_k - a_{k + 1} } = b_{k + 1} \le a_{m + 1}$
So:
:$0 \le b_{k + 1} \le a_{m + 1}$
or:
:$0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \cdots + a_{k + 1} \le a_{m + 1}$
Thus for even $k$ it follows that $\map P k \implies \map P {k + 1}$.
So for both even and odd $k$ it follows that $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.
Category:Alternating Series Test
Alternating Sum and Difference of Binomial Coefficients for Given n

== Theorem ==
:$\ds \forall n \in \Z_{\geq 0}: \sum_{i \mathop = 0}^n \paren {-1}^i \binom n i = \delta_{n 0}$
where:
:$\dbinom n i$ is a binomial coefficient
:$\delta_{n 0}$ denotes the Kronecker delta.
=== Corollary ===
== Lemma ==
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Binomial Coefficients: $3.1.7$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Properties of Binomial Coefficients: $3.8$
* : Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem: Problems $1.2$: $3 \ \text{(b)}$
* : $\S 1.2.6$: Binomial Coefficients: Exercise $36$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Properties of Binomial Coefficients: $3.8.$
Category:Alternating Sum and Difference of Binomial Coefficients for Given n
Category:Binomial Coefficients
Alternating Sum and Difference of Binomial Coefficients for Given n/Corollary

== Corollary to Alternating Sum and Difference of Binomial Coefficients for Given n ==
:$\ds \sum_{i \mathop \in \Z} \paren {-1}^i \binom n i = \delta_{n 0}$
== Proof ==
From the definition of the binomial coefficient, when $i < 0$ and $i > n$ we have $\dbinom n i = 0$.
The result follows from Alternating Sum and Difference of Binomial Coefficients for Given n.
Category:Binomial Coefficients
Category:Alternating Sum and Difference of Binomial Coefficients for Given n
Alternating Sum and Difference of Binomial Coefficients for Given n/Lemma

== Lemma for Alternating Sum and Difference of Binomial Coefficients for Given $n$ ==
:$\ds \sum_{i \mathop = 0}^0 \binom 0 0 = 1$
== Proof ==
We have by definition of vacuous summation that:
:$\ds \forall n \in \Z: n < 0: \sum_{i \mathop = 0}^n \binom n 1 = 0$
Then from Zero Choose Zero:
:$\ds \sum_{i \mathop = 0}^0 \binom 0 0 = 1$
</onlyinclude>
Category:Alternating Sum and Difference of Binomial Coefficients for Given n
Alternating Sum and Difference of Binomial Coefficients for Given n/Proof 2

== Theorem ==
== Proof ==
=== Lemma ===
Let $n > 0$.
From the Binomial Theorem, we have that:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{i \mathop = 0}^n \binom n i x^{n - i} y^i$
Putting $x = 1, y = -1$, we get:
== Sources ==
* : $\text I$. Algebra: The Binomial Theorem: Relations between coeffficients
Category:Alternating Sum and Difference of Binomial Coefficients for Given n
Alternating Sum and Difference of Binomial Coefficients for Given n/Proof 3

== Theorem ==
== Proof ==
=== Lemma ===
Let $n > 0$.
The assertion can be expressed:
:$\ds \sum_{i \mathop \le n} \paren {-1}^i \binom n i = 0$ for all $n > 0$
as $\dbinom n i = 0$ when $i < 0$ by definition of binomial coefficient.
From Alternating Sum and Difference of r Choose k up to n we have:
:$\ds \sum_{i \mathop \le n} \paren {-1}^i \binom r i = \paren {-1}^n \binom {r - 1} n$
Putting $r = n$ we have:
:$\ds \sum_{i \mathop \le n} \paren {-1}^i \binom n i = \paren {-1}^n \binom {n - 1} n$
As $n - 1 < n$ it follows from the definition of binomial coefficient that:
:$\dbinom {n - 1} n = 0$
Category:Alternating Sum and Difference of Binomial Coefficients for Given n
Alternating Sum and Difference of Sequence of Squares

== Theorem ==
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \forall n \ge 1: \sum_{j \mathop = 1}^{2 n} \paren {-1}^{j - 1} j^2 = -n \paren {2 n + 1}$
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to be true.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \sum_{j \mathop = 1}^{2 k} \paren {-1}^{j - 1} j^2 = -k \paren {2 k + 1}$
Then we need to show:
:$\ds \sum_{j \mathop = 1}^{2 \paren {k + 1} } \paren {-1}^{j - 1} j^2 = -\paren {k + 1} \paren {2 \paren {k + 1} + 1}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \ge 1: \sum_{j \mathop = 1}^{2 n} \paren {-1}^{j - 1} j^2 = -n \paren {2 n + 1}$
== Sources ==
* : $\text I$. Algebra: The Method of Induction: Exercises $\text {II}$: $4$
Category:Sums of Sequences
Category:Proofs by Induction
Alternating Sum and Difference of r Choose k up to n

== Theorem ==
Let $r \in \R, n \in \Z$.
Then:
:$\ds \sum_{k \mathop \le n} \paren {-1}^k \binom r k = \paren {-1}^n \binom {r - 1} n$
== Proof 1 ==
== Proof 2 ==
Category:Binomial Coefficients
Category:Alternating Sum and Difference of r Choose k up to n
Alternating Sum and Difference of r Choose k up to n/Proof 2

== Theorem ==
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{k \mathop \le n} \paren {-1}^k \binom r k = \paren {-1}^n \binom {r - 1} n$
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P m$ is true, where $m \ge 1$, then it logically follows that $\map P {m + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_{k \mathop \le m} \paren {-1}^k \binom r k = \paren {-1}^m \binom {r - 1} m$
from which it is to be shown that:
:$\ds \sum_{k \mathop \le m + 1} \paren {-1}^k \binom r k = \paren {-1}^{m + 1} \binom {r - 1} {m + 1}$
=== Induction Step ===
This is the induction step:
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \ds \sum_{k \mathop \le n} \paren {-1}^k \binom r k = \paren {-1}^n \binom {r - 1} n$
== Sources ==
* : $\text {3-1}$ Permutations and Combinations: Exercise $6$
Category:Alternating Sum and Difference of r Choose k up to n
Alternating Sum of Sequence of Odd Cubes over Fourth Power plus 4

== Theorem ==
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{k \mathop = 0}^n \dfrac {\paren {-1}^k \paren {2 k + 1}^3} {\paren {2 k + 1}^4 + 4} = \dfrac {\paren {-1}^n \paren {n + 1} } {4 \paren {n + 1}^2 + 1}$
=== Basis for the Induction ===
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P {r - 1}$ is true, where $r \ge 1$, then it logically follows that $\map P r$ is true.
So this is the induction hypothesis:
:$\ds \sum_{k \mathop = 0}^{r - 1} \dfrac {\paren {-1}^k \paren {2 k + 1}^3} {\paren {2 k + 1}^4 + 4} = \dfrac {\paren {-1}^r r} {4 r^2 + 1}$
from which it is to be shown that:
:$\ds \sum_{k \mathop = 0}^r \dfrac {\paren {-1}^k \paren {2 k + 1}^3} {\paren {2 k + 1}^4 + 4} = \dfrac {\paren {-1}^r \paren {r + 1} } {4 \paren {r + 1}^2 + 1}$
=== Induction Step ===
This is the induction step:
We refactorise the denominator:
and so the denominator is seen to be:
:$\paren {4 r^2 + 1}^2 \paren {4 \paren {r + 1}^2 + 1}$
Similarly, now the hard work has been done, for the numerator:
There exists a common factor of $\paren {4 r^2 + 1}$ in the numerator and the denominator, which can be cancelled, leaving:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 0}: \sum_{k \mathop = 0}^n \dfrac {\paren {-1}^k \paren {2 k + 1}^3} {\paren {2 k + 1}^4 + 4} = \dfrac {\paren {-1}^n \paren {n + 1} } {4 \paren {n + 1}^2 + 1}$
== Sources ==
* : $\S 1.2.1$: Mathematical Induction: Exercise $11$
Category:Sums of Sequences
Alternating Summation of Binomial Coefficient of Summation of Binomial Coefficient of Sequence

== Theorem ==
Let $\sequence a, \sequence b$ be real sequences which satisfy the condition:
:$a_n = \ds \sum_{r \mathop = 0}^n \binom n r b_r$
Then:
:$\ds \paren {-1}^n b_n = \sum_{s \mathop = 0}^n \binom n s \paren {-1}^s a_s$
== Proof ==
For $n - r > 1$ we can use Binomial Theorem for Integral Index with $x = 1$ and $y = -1$:
:$0 = \paren {1 - 1}^{n - r} = \ds \sum_{s \mathop = 0}^{n \mathop - r} \binom {n - r} s \paren {-1}^s$
Hence all terms of $(1)$ vanish except for where $n - r$.
That term is:
:$\paren {-1}^n b_n$
== Sources ==
* : $\text {3-1}$ Permutations and Combinations: Exercise $12$
Category:Binomial Coefficients
Alternative Definition of Ordinal

== Theorem ==
A set $S$ is an ordinal $S$ is transitive and is strictly well-ordered by the $\in$-relation.
== Proof ==
=== Necessary Condition ===
Suppose that $S$ is an ordinal.
Then $S$ is transitive.
By definition, the strict well-ordering on $S$ is given by the $\in$-relation.
Hence, the necessary condition is satisfied.
=== Sufficient Condition ===
Suppose that $S$ is a transitive set that is strictly well-ordered by the $\in$-relation.
Let $a \in S$.
Then:
That is, $S$ is an ordinal.
== Also see ==
* Alternative Definition of Ordinal in Well-Founded Theory
== Sources ==
* : $\S 7.3$, $\S 7.4$
Category:Ordinals
Alternative Definition of Ordinal in Well-Founded Theory

== Theorem ==
A set $S$ is an ordinal $S$ is transitive and $\forall x, y \in S: \paren {x \in y \lor x = y \lor y \in x}$.
== Proof ==
=== Forward Implication ===
Let $S$ be an ordinal.
By Alternative Definition of Ordinal, $S$ is transitive and strictly well-ordered by the epsilon relation.
By Strict Well-Ordering is Strict Total Ordering, $S$ is strictly totally ordered by $\in$.
Thus:
:$\forall x, y \in S: \paren {x \in y \lor x = y \lor y \in x}$
=== Reverse Implication ===
Let $S$ be a transitive set such that for any $x, y \in S$, $x \in y \lor y \in x \lor x = y$.
We first show that $\in$ is a strict ordering of $S$.
;Asymmetric: Let $x, y \in S$.
By Epsilon Relation is Strictly Well-Founded, $\set {x, y}$ has a strictly minimal element under $\in$.
Thus $x \notin t$ or $y \notin x$.
;Transitive: Let $x, y, z \in S$ with $x \in y$ and $y \in z$.
By assumption, $x = z$, $x \in z$, or $z \in x$.
 that $x = z$.
Then $x \in y$ and $y \in x$, contradicting the fact that $\Epsilon$ is asymmetric.
Suppose that $z \in x$.
Then $x \in y$, $y \in z$, and $z \in x$.
Thus the set $\set {x, y, z}$ has no strictly minimal element under $\Epsilon$, contradicting Epsilon Relation is Strictly Well-Founded.
Thus $x \in z$.
Thus $\in$ is a strict ordering of $S$.
Let $T$ be a non-empty subset of $S$.
By Epsilon Relation is Strictly Well-Founded, $T$ has a strictly minimal element $m$ under $\Epsilon$.
From Strictly Minimal Element is Minimal Element, $m$ is a minimal element of $T$ under $\Epsilon$.
From Minimal Element in Toset is Unique and Smallest, $m$ is the smallest element of $T$.
It follows by definition that $\Epsilon$ strictly well-orders $S$.
Thus by Alternative Definition of Ordinal, $S$ is an ordinal.
== Also see ==
* Alternative Definition of Ordinal
== Sources ==
* : $\S 7.3$, $\S 7.4$
Category:Ordinals
Alternative Hypothesis/Examples/Arbitrary Example 1

== Examples of Alternative Hypothesis ==
Let $\delta$ be a hypothesis test.
Let $H_0$ and $H_1$ be the null hypothesis and alternative hypothesis of $\delta$ respectively.
Let $\theta$ be the test statistic which is being used to determine whether $H_0$ or $H_1$ holds.
Let the null hypothesis state that $T = 2$:
:$H_0: \theta = 2$
The alternative hypothesis may be, for example:
:$H_1: \theta \ne 2$
or that:
:$H_1: \theta > 2$
== Sources ==
* 
* 
Category:Examples of Alternative Hypothesis
Altitude, Median and Perpendicular Bisector Coincide iff Triangle is Isosceles

== Theorem ==
Let $\triangle ABC$ be a triangle.
Then:
::the altitude from $AB$ to $C$
::the median from $AB$ to $C$
::the perpendicular bisector of $AB$
:are all the same straight line
:
:$\triangle ABC$ is isosceles where $AB$ is the base.
== Proof ==
=== Necessary Condition ===
Let $\triangle ABC$ be an isosceles triangle whose base is $AB$.
Let $D$ be the midpoint of $AB$.
:
By definition of isosceles triangle, $AC = BC$.
We have $AD = DB$ by construction, and $CD$ is common.
So by Triangle Side-Side-Side Congruence:
:$\triangle ACD = \triangle BCD$
From Two Angles on Straight Line make Two Right Angles:
:$\angle ADC + \angle BDC$ equals two right angles
and as $\angle ADC = \angle BDC$ they are each both right angles.
Thus $CD$ is the perpendicular bisector of $AB$.
By definition:
:As $D$ is the midpoint of $AB$, $CD$ is the median from $AB$ to $C$.
:As $CD$ is perpendicular to $AB$ and passes through $C$, which is a vertex of $\triangle ABC$, $CD$ is the altitude from $AB$ to $C$.
Thus all three lines coincide.
=== Converse Statement ===
Let $AC \ne BC$ in $\triangle ABC$.
Let $CD$ be the altitude from $AB$ to $C$.
From Triangle with Two Equal Angles is Isosceles, it follows that $\angle CAB \ne \angle CBA$.
Thus $\triangle ACD \ne \triangle BCD$.
Let $E$ be the midpoint of $AB$.
Let $EF$ be the perpendicular bisector of $AB$.
By definition, $CE$ is the median from $AB$ to $C$.
:
 $D = E$.
Then:
:$AD = BD$
:$AC = BC$
:$CD$ is common.
So by Triangle Side-Side-Side Congruence:
:$\triangle ACD = \triangle BCD$
But we have that $\angle CAB \ne \angle CBA$.
So by Proof by Contradiction it follows that $D \ne E$.
It follows that $CD$, $CE$ and $EF$ are all different lines.
Category:Isosceles Triangles
Category:Perpendicular Bisectors
Category:Medians of Triangles
Altitude of North Celestial Pole equals Latitude of Observer

== Theorem ==
Let $O$ be an observer of the celestial sphere.
Let $P$ be the position of the north celestial pole with respect to $O$.
Let $a$ denote the altitude of $P$.
Let $\phi$ denote the (terrestrial) latitude of $O$.
Then:
:$a = \phi$
== Proof ==
Let $z$ denote the zenith distance of $P$.
Let $\psi$ denote the (terrestrial) colatitude of $O$.
By definition we have:
:$a = 90 \degrees - z$
:$\phi - 90 \degrees - \psi$
Then:
Hence the result.
== Sources ==
* : Chapter $\text {II}$. The Celestial Sphere: $18$. Altitude and azimuth.
Category:Spherical Astronomy
Altitudes of Triangle Bisect Angles of Orthic Triangle

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $\triangle DEF$ be its orthic triangle.
The altitudes of $\triangle ABC$ are the angle bisectors of $\triangle DEF$.
== Proof ==
:
Consider the triangles $\triangle ABE$ and $\triangle ACF$.
We have that:
:$\angle FAC$ and $\angle BAE$ are common
:$\angle AFC$ and $\angle AEB$ are both right angles
and it follows from Triangles with Two Equal Angles are Similar that $\triangle ABE$ and $\triangle ACF$ are similar.
Thus:
:$\angle ABE = \angle ACF$
Consider the quadrilateral $\Box BFHD$.
We have that $\angle BFH$ and $\angle BDH$ are both right angles.
Thus two opposite angles of $\Box BFHD$ sum to two right angles
So by Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles $\Box BFHD$ is a cyclic quadrilateral.
From Angles in Same Segment of Circle are Equal:
:$\angle FBH = \angle FDH$.
By similar analysis of quadrilateral $\Box DHEC$, we note that:
:$\angle HDE = \angle HCE$
But then we have:
:$\angle FBH = \angle ABE$
and:
:$\angle HCE = \angle ACF$
Hence it follows that:
:$\angle FDH = \angle HDE$
demonstrating that $AD$ is the angle bisector of $\angle FDE$.
The same argument applies to $\angle FDE$ and $\angle FED$.
Hence the result.
== Sources ==
* 
* 
* 
* 
Category:Orthic Triangles
Altitudes of Triangle Meet at Point

== Theorem ==
Let $\triangle ABC$ be a triangle.
The altitudes of $\triangle ABC$ all intersect at the same point.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Definition:Orthocenter: the name given to that point where the altitudes intersect
== Sources ==
* : $\text {IV}$. Pure Geometry: Plane Geometry: The orthocentre
* 
* 
Category:Altitudes of Triangle Meet at Point
Category:Orthocenters of Triangles
Category:Triangles
Altitudes of Triangle Meet at Point/Proof 1

== Theorem ==
== Proof ==
From Triangle is Medial Triangle of Larger Triangle, it is possible to construct $\triangle DEF$ such that $\triangle ABC$ is the medial triangle of $\triangle DEF$.
Let $AJ$, $BG$ and $CH$ be the altitudes of $\triangle ABC$.
:
By definition of medial triangle, $A$, $B$ and $C$ are all midpoints of the sides $DF$, $DE$ and $EF$ of $\triangle DEF$ respectively.
As $DF$, $DE$ and $EF$ are parallel to $BC$, $AC$ and $AB$ respectively, $AJ$, $BG$ and $CH$ are perpendicular to $DF$, $DE$ and $EF$ respectively.
Thus, by definition, $AJ$, $BG$ and $CH$ are the perpendicular bisectors of $DF$, $DE$ and $EF$ respectively.
The result follows from Perpendicular Bisectors of Triangle Meet at Point.
Category:Altitudes of Triangle Meet at Point
Altitudes of Triangle Meet at Point/Proof 2

== Theorem ==
== Proof ==
In $\triangle ABC$ construct the altitude from vertex $A$ to side $BC$ at $P$.
Also draw the altitude from vertex $B$ to side $AC$ at $Q$.
By Two Straight Lines make Equal Opposite Angles:
:$\angle AOQ = \angle BOP$
Given:
:$\angle AQO = \angle BPO = $ one right angle.
By Triangles with Two Equal Angles are Similar:
:$\triangle AOQ \sim \triangle BOP$
and:
:$\triangle AOQ \sim \triangle ACP \sim \triangle BOP$
By the definition of similarity:
:$\dfrac {OP} {BP} = \dfrac {CP} {AP}$
Rearrange:
:$OP = \dfrac {BP \cdot CP} {AP}$
Note there is no term related to side $AC$.
If we had used side $AB$ instead of $AC$, we would have the same result.
Therefore, the altitude from vertex $C$ to side $AB$ meets both $AP$ and $BQ$ at $O$.
== Historical Note ==
== Sources ==
* : Chapter $21$: A Kind of Symmetry
Category:Altitudes of Triangle Meet at Point
Ambiguous Case for Spherical Triangle

== Theorem ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
=== Side-Side-Angle ===
=== Angle-Angle-Side ===
These are known as the ambiguous cases (for the spherical triangle).
== Also see ==
* Ambiguous Case for Triangle Side-Side-Angle Congruence
== Sources ==
* 
* 
Category:Ambiguous Case for Spherical Triangle
Category:Ambiguous Case
Category:Spherical Triangles
Category:Named Theorems