nyuuzyou/wikis huggingface dataset:
Triangle Inequality for Integrals/Real

== Theorem ==
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.
Then:
:$\ds \size {\int_a^b \map f t \rd t} \le \int_a^b \size {\map f t} \rd t$
== Proof ==
From Negative of Absolute Value, we have for all $a \in \closedint a b$:
:$-\size {\map f t} \le \map f t \le \size {\map f t}$
Thus from Relative Sizes of Definite Integrals:
:$\ds -\int_a^b \size {\map f t} \rd t \le \int_a^b \map f t \rd t \le \int_a^b \size {\map f t} \rd t$
Hence the result.
== Also see ==
* Triangle Inequality for Integrals, of which this is a special case
== Sources ==
* : $\S 13.24$
Category:Triangle Inequality for Integrals
Category:Integral Calculus
Triangle Inequality for Series/Lebesgue Spaces

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $p \in \R$, $p \ge 1$.
Let $\sequence {f_n}_{n \mathop \in \N} \in \map {\LL^p} {\mu}$ be a sequence of $p$-integrable functions, that is, a sequence in Lebesgue $p$-space.
Suppose that for all $n \in \N$, $f_n \ge 0$ holds pointwise.
Then:
:$\ds \norm {\sum_{n \mathop = 1}^\infty f_n}_p \le \sum_{n \mathop = 1}^\infty \norm {f_n}_p$
where $\norm {\, \cdot \,}_p$ denotes the $p$-seminorm.
== Proof ==
== Sources ==
* : $12.6$
Category:Series
Category:Lebesgue Spaces
Category:Triangle Inequality
Triangle Inequality for Summation over Finite Set

== Theorem ==
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f : S \to \mathbb A$ be a mapping.
Let $\size {\, \cdot\,}$ denote the standard absolute value.
Let $\size f$ be the absoute value of $f$.
Then we have the inequality of summations on finite sets:
:$\ds \size {\sum_{s \mathop \in S} \map f s} \le \sum_{s \mathop \in S} \size {\map f s}$
== Outline of Proof ==
Using the definition of summation, we reduce this to Triangle Inequality for Indexed Summations.
== Proof ==
Let $n$ be the cardinality of $S$.
Let $\sigma: \N_{< n} \to S$ be a bijection, where $\N_{<n}$ is an initial segment of the natural numbers.
By definition of summation, we have to prove the following inequality of indexed summations:
:$\ds \size {\sum_{i \mathop = 0}^{n - 1} \map f {\map \sigma i} } \le \sum_{i \mathop = 0}^{n - 1} \map {\paren {\size f \circ \sigma} } i$
By Absolute Value of Mapping Composed with Mapping:
:$\size f \circ \sigma = \size {f \circ \sigma}$
The above equality now follows from Triangle Inequality for Indexed Summations.
== Also see ==
* Triangle Inequality for Series
Category:Summations
Category:Triangle Inequality
Triangle Inequality for Variation of Complex Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be measurable space.
Let $\mu$ and $\nu$ be two complex measures.
Let $\cmod {\mu + \nu}$, $\cmod \mu$ and $\cmod \nu$ be the variation of $\mu + \nu$, $\mu$ and $\nu$ respectively.
Then:
:$\cmod {\mu + \nu} \le \cmod \mu + \cmod \nu$ 
== Proof ==
Let $\map P A$ be the set of finite partitions of $A$ into $\Sigma$-measurable sets.
Let $\set {A_1, A_2, \ldots, A_n} \in \map P A$. 
Then from Triangle Inequality for Complex Numbers we have:
:$\ds \sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} + \map \nu {A_j} } \le \sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } + \sum_{j \mathop = 1}^n \cmod {\map \nu {A_j} } \le \map {\cmod \mu} A + \map {\cmod \nu} A$
by the definition of variation.
Taking the supremum over $\set {A_1, A_2, \ldots, A_n} \in \map P A$ we have:
:$\map {\cmod {\mu + \nu} } A \le \map {\cmod \mu} A + \map {\cmod \nu} A$ 
Category:Variation of Complex Measure
Category:Triangle Inequality
Triangle Inequality on Distance from Point to Subset

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$.
Then:
:$\forall x, y \in A: \map d {x, H} \le \map d {x, y} + \map d {y, H}$
where $\map d {x, H}$ denotes the distance between $x$ and $H$.
== Proof ==
Hence the result.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Exercise $6$
* : $4$: The Hausdorff condition: Exercise $4.3: 5 \ \text {(b)}$
Category:Distance Function
Category:Triangle Inequality
Triangle Right-Angle-Hypotenuse-Side Congruence

== Theorem ==
If two right triangles have:
:their hypotenuses equal
:another of their respective sides equal
they will also have:
:their third sides equal
:the remaining two angles equal to their respective remaining angles.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.15$
* $(4)$
* $(4)$
Category:Triangle Right-Angle-Hypotenuse-Side Congruence
Category:Right Triangles
Category:Congruence (Geometry)
Triangle Right-Angle-Hypotenuse-Side Congruence/Also known as

== Triangle Right-Angle-Hypotenuse-Side Congruence: Also known as ==
Triangle Right-Angle-Hypotenuse-Side Congruence is also known as RHS or the RHS Condition.
However, because RHS is also used as a standard abbreviation for the of an equation, it is deprecated on on account of ambiguity.
== Sources ==
* $(4)$
* $(4)$
Category:Triangle Right-Angle-Hypotenuse-Side Congruence
Triangle Right-Angle-Hypotenuse-Side Congruence/Proof 1

== Theorem ==
== Proof ==
Let $\triangle ABC$ and $\triangle DEF$ be two triangles having sides $AB = DE$ and $AC = DF$, and with $\angle ABC = \angle DEF = 90^\circ$.
By Pythagoras' Theorem:
:$BC = \sqrt {AB^2 + AC^2}$
and:
:$EF = \sqrt {DE^2 + DF^2}$
:$\therefore BC = \sqrt {AB^2 + AC^2} = \sqrt {DE^2 + DF^2} = EF$
The part that the remaining two angles are equal to their respective remaining angles follows from Triangle Side-Side-Side Congruence.
Category:Triangle Right-Angle-Hypotenuse-Side Congruence
Triangle Right-Angle-Hypotenuse-Side Congruence/Proof 2

== Theorem ==
== Proof ==
Let $\triangle ADB$ and $\triangle ADC$ both be right triangles.
Let them have equal hypotenuse and one leg ($AD$) equal.
:
:$AB = AC$
:$\angle ADB = \angle ADC = \ $ one right angle
:$AD$ is shared
So the two triangles can be drawn as shown with $BD$ and $DC$ joined at $D$.
By addition:
:$\angle BDA + \angle CDA = \angle BDC = \ $ two right angles
By Two Angles making Two Right Angles make Straight Line:
:$BDC$ is one straight line
:the points $BDC$ are collinear
So:
:$\triangle ABC$ is a triangle.
By definition of isosceles triangle:
:$\triangle ABC$ is isosceles.
Since $AD \perp BDC$, by definition of perpendicular bisector:
:$BD = DC$
By Triangle Side-Angle-Side Congruence:
:$\triangle ABD \cong \triangle ACD$
Category:Triangle Right-Angle-Hypotenuse-Side Congruence
Triangle Side-Angle-Angle Congruence

== Theorem ==
If two triangles have:
:two angles equal to two angles, respectively
:the sides opposite one pair of equal angles equal
then the remaining angles are equal, and the remaining sides equal the respective sides.
That is to say, if two pairs of angles and a pair of opposite sides are equal, then the triangles are congruent.
== Proof ==
:
Let:
:$\angle ABC = \angle DEF$
:$\angle BCA = \angle EFD$
:$AB = DE$
 that $BC \ne EF$.
If this is the case, one of the two must be greater.
, let $BC > EF$.
We construct a point $H$ on $BC$ such that $BH = EF$, and then we construct the segment $AH$.
Now, since we have:
:$BH = EF$
:$\angle ABH = \angle DEF$
:$AB = DE$
from Triangle Side-Angle-Side Congruence we have:
:$\angle BHA = \angle EFD$
But from External Angle of Triangle is Greater than Internal Opposite, we have:
:$\angle BHA > \angle HCA = \angle EFD$
which is a contradiction.
Therefore $BC = EF$.
So from Triangle Side-Angle-Side Congruence:
:$\triangle ABC = \triangle DEF$
== Also known as ==
== Also see ==
* Triangle Angle-Side-Angle Congruence
* Triangle Side-Side-Side Congruence
== Sources ==
* : Book $\text{I}$. Propositions
* $(3)$
* $(3)$
Category:Triangle Side-Angle-Angle Congruence
Category:Triangles
Category:Congruence (Geometry)
Triangle Side-Angle-Angle Congruence/Also known as

== Triangle Side-Angle-Angle Congruence: Also known as ==
Triangle Side-Angle-Angle Congruence is also known as:
:Triangle Angle-Angle-Side Congruence
:SAA or the SAA Condition
:AAS or the AAS Condition
== Sources ==
* $(3)$
* $(3)$
Category:Triangle Side-Angle-Angle Congruence
Triangle Side-Angle-Side Congruence

== Theorem ==
If $2$ triangles have:
:$2$ sides equal to $2$ sides respectively
:the angles contained by the equal straight lines equal
they will also have:
:their third sides equal
:the remaining two angles equal to their respective remaining angles, namely, those which the equal sides subtend.
== Proof ==
Let $\triangle ABC$ and $\triangle DEF$ be $2$ triangles having sides $AB = DE$ and $AC = DF$, and with $\angle BAC = \angle EDF$.
If $\triangle ABC$ is placed on $\triangle DEF$ such that:
:the point $A$ is placed on point $D$, and 
:the line $AB$ is placed on line $DE$
then the point $B$ will also coincide with point $E$ because $AB = DE$.
So, with $AB$ coinciding with $DE$, the line $AC$ will coincide with the line $DF$ because $\angle BAC = \angle EDF$.
Hence the point $C$ will also coincide with the point $F$, because $AC = DF$.
But $B$ also coincided with $E$.
Hence the line $BC$ will coincide with line $EF$.
(Otherwise, when $B$ coincides with $E$ and $C$ with $F$, the line $BC$ will not coincide with line $EF$ and two straight lines will enclose a region which is impossible.)
Therefore $BC$ will coincide with $EF$ and be equal to it.
Thus the whole $\triangle ABC$ will coincide with the whole $\triangle DEF$ and thus $\triangle ABC = \triangle DEF$.
The remaining angles on $\triangle ABC$ will coincide with the remaining angles on $\triangle DEF$ and be equal to them.
== Also known as ==
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.10$
* $(2)$
* $(2)$
Category:Triangle Side-Angle-Side Congruence
Category:Triangles
Category:Congruence (Geometry)
Triangle Side-Side-Side Congruence

== Theorem ==
Let two triangles have all $3$ sides equal.
Then they also have all $3$ angles equal.
Thus two triangles whose sides are all equal are themselves congruent.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.14$
* $(1)$
* $(1)$
* : $\S 1$: What Is Curvature? The Euclidean Plane
Category:Triangle Side-Side-Side Congruence
Category:Triangles
Category:Congruence (Geometry)
Triangle Side-Side-Side Congruence/Proof 1

== Theorem ==
== Proof ==
:
Let $\triangle ABC$ and $\triangle DEF$ be two triangles such that:
:$AB = DE$
:$AC = DF$
:$BC = EF$
Suppose $\triangle ABC$ were superimposed over $\triangle DEF$ so that point $B$ is placed on point $E$ and the side $BC$ on $EF$.
Then $C$ will coincide with $F$, as $BC = EF$ and so $BC$ coincides with $EF$.
 $BA$ does not coincide with $ED$ and $AC$ does not coincide with $DF$.
Then they will fall as, for example, $EG$ and $GF$.
Thus there will be two pairs of straight line segments constructed on the same line segment, on the same side as it, meeting at different points.
But from , these points are coincident.
It follows by Proof by Contradiction that:
:$BA$ coincides with $ED$
:$AC$ coincides with $DF$.
Therefore $\angle BAC$ coincides with $\angle EDF$ and therefore:
:$\angle BAC = \angle EDF$
The same argument can be applied to the other two sides.
Thus we have shown that all corresponding angles are equal.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Triangle Side-Side-Side Congruence
Triangle Side-Side-Side Congruence/Proof 2

== Theorem ==
== Proof ==
Let $\triangle ABC$ and $\triangle DEF$ have all three pairs of corresponding sides equal.
Let $AC$ be the longest side of $\triangle ABC$. 
If there is more than one longest side, choose an arbitrary one.
We have :
:$AC = DF$
There are two cases:
case 1
The two triangles can be arranged as shown. They are reflections.
:
case 2
Otherwise, they are superimposable by the method in Triangle Side-Angle-Side Congruence).
In the words of Euclid (Euclid:Proposition/I/4:
:"If the triangle ABC is superposed on the triangle DEF, and if the point A is placed on the point D and the straight line AB on DE, then the point B also coincides with E, because AB equals DE..."
If so, then reflect $\triangle DEF$ in $DF$.
Now arrange $\triangle ABC$ and $\triangle DEF$ with $AC = DF$ shared, as shown in the figure.
Construct $BE$.
We have :
:$AB = AE$
:$BC = EC$
By the definition of isosceles triangles:
:$\triangle ABE$ and $\triangle BEC$ are both isosceles.
By Isosceles Triangle has Two Equal Angles:
:$\angle ABE = \angle AEB$
:$\angle EBC = \angle BEC$
By addition:
:$\angle ABC = \angle AEC$
By Triangle Side-Angle-Side Congruence:
:$\triangle ABC \cong \triangle DEF$
== Sources ==
* 
Category:Triangle Side-Side-Side Congruence
Triangle is Convex Set

== Theorem ==
The interior of a triangle embedded in $\R^2$ is a convex set.
== Proof ==
Denote the triangle as $\triangle$, and the interior of the boundary of $\triangle$ as $\Int \triangle$.
From Boundary of Polygon is Jordan Curve, it follows that the boundary of $\triangle$ is equal to the image of a Jordan curve, so $\Int \triangle$ is well-defined.
Denote the vertices of $\triangle$ as $A_1, A_2, A_3$.
For $i \in \set {1, 2, 3}$, put $j = i \bmod 3 + 1$, $k = \paren {i + 1} \bmod 3 + 1$, and:
:$U_i = \set {A_i + s t \paren {A_j - A_i} + \paren {1 - s} t \paren {A_k - A_i}: s \in \openint 0 1, t \in \R_{>0} }$
Suppose that the angle $\angle A_i$ between $A_j - A_i$ and $A_k - A_i$ is non-convex.
As $\angle A_i$ is an internal angle in $\triangle$, it follows from definition of polygon that $\angle A_i$ cannot be zero or straight.
Then $\angle A_i$ is larger than a straight angle, which is impossible by Sum of Angles of Triangle Equals Two Right Angles.
It follows that $\angle A_i$ is convex.
From Characterization of Interior of Triangle, it follows that:
:$\ds \Int \triangle = \bigcap_{i \mathop = 1}^3 U_i$
From Interior of Convex Angle is Convex Set, it follows for $i \in \set {1, 2, 3}$ that $U_i$ is a convex set.
The result now follows from Intersection of Convex Sets is Convex Set (Vector Spaces).
Category:Convex Sets (Vector Spaces)
Triangle is Medial Triangle of Larger Triangle

== Theorem ==
Let $\triangle ABC$ be a triangle.
$\triangle ABC$ is the medial triangle of a larger triangle.
== Construction ==
Consider $\triangle ABC$.
Let $D, E, F$ be constructed so that:
:$DE \parallel AC$
:$EF \parallel AB$
:$DF \parallel BC$
Then $\triangle ABC$ is the medial triangle of $\triangle DEF$.
== Proof ==
By , it is possible to construct exactly one straight line parallel to each of $AC$, $BC$ and $AC$.
:
From Parallelism implies Equal Corresponding Angles:
:$\angle ABC = \angle ECB = \angle DAB$
:$\angle ACB = \angle CBE = \angle CAF$
:$\angle CAB = \angle ABD = \angle ACF$
By Triangle Angle-Side-Angle Congruence:
:$\triangle ABC = \triangle ABD$
:$\triangle ABC = \triangle CFA$
:$\triangle ABC = \triangle ECB$
Thus:
:$FC = CE$
:$DB = BE$
:$FA = AD$
The result follows by definition of medial triangle.
== Sources ==
* 
Category:Medial Triangles
Triangle is Orthic Triangle of Triangle formed from Excenters

== Theorem ==
Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Let $I$ be the incenter of $\triangle ABC$.
Let $I_a$, $I_b$ and $I_c$ be the excenters of $\triangle ABC$ $a$, $b$ and $c$ respectively.
Let $\triangle I_a I_b I_c$ be the triangle whose vertices are those excenters of $I_a$, $I_b$ and $I_c$.
Then:
:$\triangle ABC$ is the orthic triangle of $\triangle I_a I_b I_c$
and:
:$I$ is the orthocenter of $\triangle I_a I_b I_c$.
== Proof ==
:
From Construction of Excircle to Triangle, it is seen that:
:$A I_b$ is the angle bisector of $\angle PAC$
:$A I_c$ is the angle bisector of $\angle QAB$.
Hence $I_b A I_c$ is a straight line.
From the construction in Excenters and Incenter of Orthic Triangle, $A I I_a$ is a straight line which is perpendicular to $I_b A I_c$.
The same argument is used to show that:
:$B I I_b$ is a straight line which is perpendicular to the straight line $I_a B I_c$
and
:$C I I_c$ is a straight line which is perpendicular to the straight line $I_a C I_b$.
It follows by definition that :
:$\triangle ABC$ is the orthic triangle of $\triangle I_a I_b I_c$
and:
:$I$ is the orthocenter of $\triangle I_a I_b I_c$.
== Sources ==
* : $\text V$. Trigonometry: The ex-circles
Category:Incircles of Triangles
Category:Excircles of Triangles
Category:Orthic Triangles
Triangle is Self-Conjugate to Unique Circle

== Theorem ==
Let $\triangle PQR$ be a triangle.
Then there exists exactly one circle $\CC$ such that $\triangle PQR$ is self-conjugate with respect to $\CC$.
== Proof ==
== Sources ==
* : Chapter $\text {III}$. The Circle: $9$. Conjugate triangles: Example $2$
Category:Conjugate Triangles
Triangle of Forces

== Theorem ==
Let $3$ forces act on a particle.
Let the magnitude and direction of these forces be represented by vectors in space such that the magnitude of the forces is proportional to the length of the vectors.
Then the forces are in equilibrium the vectors form a triangle.
This is known as a triangle of forces.
== Proof ==
== Also see ==
* Polygon of Forces
== Sources ==
* 
Category:Triangle of Forces
Category:Equilibrium (Mechanics)
Category:Force
Triangle with Two Equal Angles is Isosceles

== Theorem ==
If a triangle has two angles equal to each other, the sides which subtend the equal angles will also be equal to one another.
Hence, by definition, such a triangle will be isosceles.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.13$
Category:Isosceles Triangles
Category:Triangle with Two Equal Angles is Isosceles
Triangle with Two Equal Angles is Isosceles/Proof 1

== Theorem ==
== Proof ==
:
Let $\triangle ABC$ be a triangle in which $\angle ABC = \angle ACB$.
Suppose side $AB$ is not equal to side $AC$. Then one of them will be greater.
, Suppose $AB > AC$.
We cut off from $AB$ a length $DB$ equal to $AC$.
We draw the line segment $CD$.
Since $DB = AC$, and $BC$ is common, the two sides $DB, BC$ are equal to $AC, CB$ respectively.
Also, $\angle DBC = \angle ACB$.
So by Triangle Side-Angle-Side Congruence:
:$\triangle DBC = \triangle ACB$
But $\triangle DBC$ is smaller than $\triangle ACB$, which is absurd.
Therefore, have $AB \le AC$. 
A similar argument shows the converse, and hence $AB = AC$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Triangle with Two Equal Angles is Isosceles
Triangles with Equal Base and Same Height have Equal Area

== Theorem ==
== Proof ==
:
Let $ABC$ and $DEF$ be triangles which have equal bases $BC$ and in the same parallels $AD$ and $BF$.
Let $AD$ be produced in the directions of $G$ and $H$.
Let $BG$ through $B$ be drawn parallel to $CA$.
Let $FH$ through $F$ be drawn parallel to $DE$.
Then each of $GBCA$ and $DEFH$ are parallelograms, and by Parallelograms with Same Base and Same Height have Equal Area they have equal areas.
From Opposite Sides and Angles of Parallelogram are Equal, $ABC$ is half of $GBCA$ as $AB$ bisects it.
For the same reason, $DEF$ is half of $DEFH$.
But by Common Notion 1, $\triangle ABC = \triangle DEF$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Areas of Triangles
Triangles with Integer Area and Integer Sides in Arithmetical Sequence

== Theorem ==
The triangles with the following sides in arithmetic sequence have integer areas:
:$3, 4, 5$
:$13, 14, 15$
:$15, 28, 41$
:$15, 26, 37$
Their areas are:
:$6, 84, 126, 156$
== Proof ==
From Heron's Formula, the area $A$ of $\triangle ABC$ is given by:
:$A = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where $s = \dfrac{a + b + c} 2$ is the semiperimeter of $\triangle ABC$.
For $3, 4, 5$:
For $13, 14, 15$:
For $15, 28, 41$:
For $15, 26, 37$:
== Sources ==
* 
* : $126$
Category:Areas of Triangles
Triangles with Integer Area and Integer Sides in Arithmetical Sequence/Mistake

== Source Work ==
:
:The Dictionary
::$126$
== Mistake ==
:[$126$ is] the area of the third-smallest triangle with integral sides in arithmetical sequence and integral area: the sides are $15, 28, 41$. The first two are $3, 4, 5$ and $13, 14, 15$; the next is $15, 26, 37$, with area $156$.
Exactly what this statement is supposed to say is not clear, but the triangle with integral sides $6, 8, 10$, in arithmetic sequence, has area $24$, which is smaller than the $13, 14, 15$ triangle.
== Sources ==
* 
* : $126$
Category:Mistakes/Curious and Interesting Numbers
Triangles with One Equal Angle and Two Other Sides Proportional are Similar

== Theorem ==
Let two triangles be such that one of the angles of one triangle equals one of the angles of the other.
Let two corresponding sides which are adjacent to one of the other angles, be proportional.
Let the third angle in both triangles be either both acute or both not acute.
Then all of the corresponding angles of these triangles are equal.
Thus, by definition, such triangles are similar.
== Proof ==
Let $\triangle ABC, \triangle DEF$ be triangles such that:
:$\angle BAC = \angle EDF$
:two sides adjacent to $\angle ABC$ and $\angle DEF$ proportional, so that $AB : BC = DE : EF$
:$\angle ACB$ and $\angle DFE$ either both acute or both not acute.
We need to show that $\angle ABC = \angle DEF$ and $\angle BCA = \angle EFD$.
First assume that $\angle ACB$ and $\angle DFE$ are both acute.
:
Suppose $\angle ABC \ne \angle DEF$.
Then one of them is greater.
, let $\angle ABC > \angle DEF$.
On $AB$ and at point $B$, let $\angle ABG$ be constructed equal to $\angle DEF$.
Since $\angle BAG = \angle EDF$ and $\angle ABG = \angle DEF$, from Sum of Angles of Triangle Equals Two Right Angles:
:$\angle AGB = \angle DFE$
So $\triangle ABG$ is equiangular with $\triangle DEF$.
So from Equiangular Triangles are Similar:
:$AB : BG = DE : EF$
But :
:$DE : EF = AB : BC$
Thus from Equality of Ratios is Transitive:
:$AB : BG = AB : BC$
So from Magnitudes with Same Ratios are Equal:
:$BC = BG$
So from Isosceles Triangle has Two Equal Angles:
:$\angle BCG = \angle BGC$
We have that $\angle BCG$ is less than a right angle.
So $\angle BGC$ is less than a right angle.
So from Two Angles on Straight Line make Two Right Angles $\angle AGB$ is greater than a right angle.
But this was proved equal to $\angle DFE$.
But $\angle DFE$ is less than a right angle.
From this contradiction we conclude that:
:$\angle ABC = \angle DEF$
We also have:
:$\angle BAC = \angle DEF$
So from Sum of Angles of Triangle Equals Two Right Angles:
:$\angle ACB = \angle DFE$
So, as we wanted to show, $\triangle ABC$ is equiangular with $\triangle DEF$.
Now, suppose $\angle ACB$ and $\angle DFE$ are both not acute:
:
From the same construction we show similarly that $BC = BG$.
So, from Isosceles Triangle has Two Equal Angles:
:$\angle BCG = \angle BGC$
But $\angle BCG$ is not less than a right angle.
So neither is $\angle BGC$ less than a right angle.
So in $\triangle BGC$, $\angle BCG + \angle BGC$ are not less than two right angles.
From Two Angles of Triangle are Less than Two Right Angles, this is impossible.
So from this contradiction we conclude that:
:$\angle ABC = \angle DEF$
Hence, similarly to above, $\triangle ABC$ is equiangular with $\triangle DEF$.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Triangles