avewright/proofwiki-math huggingface dataset:
Theorem: Lines Joining Equal and Parallel Straight Lines are Parallel

The straight lines joining equal and parallel straight lines at their endpoints, in the same direction, are themselves equal and parallel.
{{:Euclid:Proposition/I/33}}

Proof:
:300px
Let $AB, CD$ be equal and parallel.
Let $AC, BD$ join their endpoints in the same direction.
Draw $BC$.
From Parallelism implies Equal Alternate Angles:
:$\angle ABC = \angle BCD$
We have that $AB, BC$ are equal to $DC, CB$ and $\angle ABC = \angle BCD$.
It follows from Triangle Side-Angle-Side Congruence that $AC = BD$.
Also, $\triangle ABC = \triangle DCB$, and thus $\angle ACB = \angle CBD$.
We have that $BC$ falling on the two straight lines $AC, BD$ makes the alternate angles equal.
Therefore from Equal Alternate Angles implies Parallel Lines:
:$AC \parallel BD$
{{qed}}
{{Euclid Note|33|I}}
Theorem: Opposite Sides and Angles of Parallelogram are Equal

The opposite sides and angles of a parallelogram are equal to one another, and either of its diameters bisects its area.
{{:Euclid:Proposition/I/34}}

Proof:
:300px
Let $ACDB$ be a parallelogram, and let $BC$ be a diameter.
By definition of parallelogram, $AB \parallel CD$, and $BC$ intersects both.
So by Parallelism implies Equal Alternate Angles:
:$\angle ABC = \angle BCD$
Similarly, by definition of parallelogram, $AC \parallel BD$, and $BC$ intersects both.
So by Parallelism implies Equal Alternate Angles:
:$\angle ACB = \angle CBD$
So $\triangle ABC$ and $\triangle DCB$ have two angles equal, and the side $BC$ in common.
So by Triangle Angle-Side-Angle Congruence:
: $\triangle ABC = \triangle DCB$
So $AC = BD$ and $AB = CD$.
Also, we have that $\angle BAC = \angle BDC$.
So we have $\angle ACB = \angle CBD$ and $\angle ABC = \angle BCD$.
So by {{EuclidCommonNotionLink|2}}:
:$\angle ACB + \angle BCD = \angle ABC + \angle CBD$
So $\angle ACD = \angle ABD$.
So we have shown that opposite sides and angles are equal to each other.
Now note that $AB = CD$, and $BC$ is common, and $\angle ABC = \angle BCD$.
So by Triangle Side-Angle-Side Congruence:
:$\triangle ABC = \triangle BCD$
So $BC$ bisects the parallelogram.
Similarly, $AD$ also bisects the parallelogram.
{{qed}}
{{Euclid Note|34|I|The use of Triangle Side-Angle-Side Congruence in this proof seems to be superfluous as the triangles were already shown to be equal using Triangle Angle-Side-Angle Congruence.  However, Euclid included the step in his proof, so the line is included here.
Note that in at least some translations of {{BookLink|The Elements|Euclid}}, the Triangle Side-Angle-Side Congruence proposition includes the extra conclusion that the two triangles themselves are equal whereas the others do not explicitly state this, but since Triangle Side-Angle-Side Congruence is used to prove the other congruence theorems, this conclusion would seem to be follow trivially in those cases.}}
Theorem: Equal Sized Triangles on Same Base have Same Height

Triangles of equal area which are on the same base, and on the same side of it, are also in the same parallels.
{{:Euclid:Proposition/I/39}}

Proof:
:300px
Let $ABC$ and $DBC$ be equal-area triangles which are on the same base $BC$ and on the same side as it.
Let $AD$ be joined.
Suppose $AD$ were not parallel to $BC$.
Then, by Construction of Parallel Line we draw $AE$ parallel to $BC$.
So by Triangles with Same Base and Same Height have Equal Area:
: $\triangle ABC = \triangle EBC$
But $\triangle ABC = \triangle DBC$, which means:
: $\triangle DBC = \triangle EBC$
But $\triangle DBC$ is bigger than $\triangle EBC$.
From this contradiction we deduce that $AE$ can not be parallel to $BC$.
In a similar way, we prove that no other line except $AD$ can be parallel to $BC$.
{{qed}}
{{Euclid Note|39|I|It is the partial converse of {{EuclidPropLink|prop = 37|title = Triangles with Same Base and Same Height have Equal Area}}.}}
Theorem: Equal Sized Triangles on Equal Base have Same Height

Triangles of equal area which are on equal bases, and on the same side of it, are also in the same parallels.
{{:Euclid:Proposition/I/40}}

Proof:
:300px
Let $ABC$ and $CDE$ be equal-area triangles which are on equal bases $BC$ and $CD$, and on the same side.
Let $AE$ be joined.
Suppose $AE$ were not parallel to $BC$.
Then, by Construction of Parallel Line we draw $AF$ parallel to $BD$.
So by Triangles with Equal Base and Same Height have Equal Area, $\triangle ABC = \triangle FCD$.
But $\triangle ABC = \triangle DCE$, which means $\triangle FCD = \triangle DCE$.
But $\triangle DCE$ is bigger than $\triangle FCD$.
From this contradiction we deduce that $AF$ can not be parallel to $BD$.
In a similar way, we prove that no other line except $AE$ can be parallel to $BD$.
{{qed}}
{{Euclid Note|40|I|It is the (partial) converse of {{EuclidPropLink|prop = 38|title = Triangles with Same Base and Same Height have Equal Area}}.<br><br>It is also apparent from the original manuscript that this proposition was a later addition by an editor who believed that there should be a proposition related to {{EuclidPropLink|prop = 39|title = Equal Sized Triangles on Same Base have Same Height}} in the same way that {{EuclidPropLink|prop = 38|title = Triangles with Equal Base and Same Height have Equal Area}} is related to {{EuclidPropLink|prop = 37|title = Triangles with Same Base and Same Height have Equal Area}}, and so on.}}
Theorem: Parallelogram on Same Base as Triangle has Twice its Area

A parallelogram on the same base as a triangle, and in the same parallels, has twice the area of the triangle.
{{:Euclid:Proposition/I/41}}

Proof:
:300px
Let $ABCD$ be a parallelogram on the same base $BC$ as a triangle $EBC$, between the same parallels $BC$ and $AE$.
Join $AC$.
Then $\triangle ABC = \triangle EBC$ from Triangles with Same Base and Same Height have Equal Area.
But from Opposite Sides and Angles of Parallelogram are Equal, $AC$ bisects $ABCD$.
So the area of parallelogram $ABCD$ is twice the area of triangle $EBC$.
{{Qed}}
{{Euclid Note|41|I}}
Theorem: Construction of Parallelogram equal to Triangle in Given Angle

A parallelogram can be constructed in a given angle the same size as any given triangle.
{{:Euclid:Proposition/I/42}}

Proof:
:400px
Let $ABC$ be the given triangle, and $D$ the given angle.
Bisect $BC$ at $E$, and join $AE$.
Construct $AG$ parallel to $EC$.
Construct $\angle CEF$ equal to $\angle D$.
Construct $CG$ parallel to $EF$.
Then $FEGC$ is a parallelogram.
Since $BE = EC$, from Triangles with Equal Base and Same Height have Equal Area, $\triangle ABE = \triangle AEC$.
So $\triangle ABC$ is twice the area of $\triangle AEC$.
But from Parallelogram on Same Base as Triangle has Twice its Area, $FECG$ is also twice the area of $\triangle AEC$.
So $FECG$ has the same area as $\triangle ABC$, and has the given angle $D$.
{{qed}}
{{Euclid Note|42|I}}
Theorem: Construction of Parallelogram in Given Angle equal to Given Polygon

A parallelogram can be constructed in a given angle the same size as any given polygon.
{{:Euclid:Proposition/I/45}}

Proof:
:600px
Let $ABCD$ be the given polygon, and let $E$ be the given angle.
Join $DB$, and construct the parallelogram $FGHK$ equal in size to $\triangle ABD$, in $\angle HKF = \angle E$.
Then construct the parallelogram $GLMH$ equal in area to $\triangle BCD$ on the line segment $GH$, in $\angle GHM = \angle E$.
We now need to show that $KFLM$ is the required parallelogram.
By {{EuclidCommonNotionLink|1}}, $\angle HKF = \angle GHM$ as both are equal to $\angle E$.
Add $\angle KHG$ to each, so as to make $\angle FKH + \angle KHG = \angle KHG + \angle GHM$.
From Parallelism implies Supplementary Interior Angles:
:$\angle FKH + \angle KHG$
Therefore $\angle KHG + \angle GHM$ equal two right angles.
So from Two Angles making Two Right Angles make Straight Line, $KH$ is in a straight line with $HM$.
From Parallelism implies Equal Alternate Angles:
:$\angle MHG = \angle HGF$
Add $\angle HGL$ to each, so as to make $\angle MHG + \angle HGL = \angle HGF + \angle HGL$.
From Parallelism implies Supplementary Interior Angles:
:$\angle MHG + \angle HGL$
Therefore $\angle HGF + \angle HGL$ equal two right angles.
So from Two Angles making Two Right Angles make Straight Line, $FG$ is in a straight line with $GL$.
From Parallelism is Transitive Relation, as $KF \parallel HG$ and $HG \parallel ML$, it follows that $KF \parallel ML$.
Similarly, from common notion 1, $KF = ML$.
As $KM$ and $FL$ join them at their endpoints, $KM \parallel FL$ and $KM = FL$ from Lines Joining Equal and Parallel Straight Lines are Parallel.
Therefore $KFLM$ is a parallelogram.
But the area of $KFLM$ equals the combined areas of $FGHK$ and $GLMH$, which are equal to the combined areas of $\triangle ABD$ and $\triangle BCD$.
Therefore from {{EuclidCommonNotionLink|2}}, $KFLM$ has the same area as the polygon $ABCD$, in the angle $E$
{{qed}}
{{Euclid Note|45|I|Note that this technique can be expanded for a polygon with any number of sides, merely by dividing the polygon up into as many triangles as it takes.}}
Theorem: Square of Sum

:$\forall x, y \in \R: \paren {x + y}^2 = x^2 + 2 x y + y^2$

Proof:
Follows directly and trivially from Square of Sum:
{{begin-eqn}}
{{eqn | l = \paren {x + y}^2
      | r = x^2 + 2 x y + y^2
      | c = Square of Sum
}}
{{eqn | ll= \leadsto
      | l = \paren {x + y}^2 - x^2
      | r = \paren {2 x + y} y
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Square of Sum

:$\forall x, y \in \R: \paren {x + y}^2 = x^2 + 2 x y + y^2$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {x + y}^2 - x^2
      | r = \paren {x + y + x} \paren {x + y - x}
      | c = Difference of Two Squares
}}
{{eqn | r = \paren {2 x + y} y
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Square of Sum

:$\forall x, y \in \R: \paren {x + y}^2 = x^2 + 2 x y + y^2$

Proof:
{{:Euclid:Proposition/II/6}}
Let $AB$ be bisected at $C$, and let $BD$ be added to it in a straight line.
Then the rectangle contained by $AD$ and $BD$ together with the square on $BC$ equals the square on $CD$.
:400px
The proof is as follows.
Construct the square $CEFD$ on $CD$, and join $DE$.
Construct $BG$ parallel to $CE$ through $B$, and let $BG$ cross $DE$ at $H$.
Construct $KM$ parallel to $AD$ through $H$.
Construct $AK$ parallel to $DF$ through $A$.
As $AC = CB$, from Parallelograms with Equal Base and Same Height have Equal Area we have that $\Box ACLK = \Box CBHL$.
From Complements of Parallelograms are Equal, $\Box CBHL = \Box HMFG$.
So $\Box ACLK = \Box HMFG$.
Add $\Box CDML$ to each.
So the whole of $\Box ADMK$ is equal to the gnomon $CDFGHL$.
But $\Box ADMK$ is the rectangle contained by $AD$ and $BD$, because $BD = DM$.
So the gnomon $CDFGHL$ is equal in area to the rectangle contained by $AD$ and $BD$.
Now $\Box LHGE$ is equal to the square on $BC$.
Add $\Box LHGE$ to each of the gnomon $CDFGHL$ and $\Box ADMK$.
Then the gnomon $CDFGHL$ together with $\Box LHGE$ equals the rectangle contained by $AD$ and $BD$ and the square on $BC$.
But the gnomon $CDFGHL$ together with $\Box LHGE$ is the square $CDFE$.
Hence the result.
{{qed}}
Theorem: Square of Sum

:$\forall x, y \in \R: \paren {x + y}^2 = x^2 + 2 x y + y^2$

Proof:
Follows from the distribution of multiplication over addition:
{{begin-eqn}}
{{eqn | l = \paren {a + 2 b}^2
      | r = \paren {a + 2 b} \cdot \paren {a + 2 b}
      | c = 
}}
{{eqn | r = a \cdot \paren {a + 2 b} + 2 b \cdot \paren {a + 2 b}
      | c = Real Multiplication Distributes over Addition
}}
{{eqn | r = a \cdot a + a \cdot 2 b + 2 b \cdot a + 2 b \cdot 2 b
      | c = Real Multiplication Distributes over Addition
}}
{{eqn | r = a^2 + 4 a b + 4 b^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Square of Sum

:$\forall x, y \in \R: \paren {x + y}^2 = x^2 + 2 x y + y^2$

Proof:
A direct application of the Binomial Theorem:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$
putting $n = 2, x = a, y = 2 b$.
{{qed}}
Theorem: Square of Sum

:$\forall x, y \in \R: \paren {x + y}^2 = x^2 + 2 x y + y^2$

Proof:
{{:Euclid:Proposition/II/8}}
That is: $4 \paren {a + b} b + a^2 = \paren {a + 2 b}^2$.
:400px
Let the straight line $AB$ be cut at random at $C$.
Then four times the rectangle contained by $AB$ and $BC$ together with the  square on $AC$ equals the square on $AB$ and $BC$ as a single straight line.
The proof is as follows.
Produce $AB$ to $D$ where $BD = BC$.
Construct the square $AEFD$ on $AD$ and join $DE$.
Draw the given figure above, in the same manner as in Square of Sum.
By Parallelograms with Equal Base and Same Height have Equal Area:
: $\Box CBKG = \Box BDNK$
: $\Box KGQR = \Box KNPR$
From Complements of Parallelograms are Equal:
: $\Box CBKG = \Box KNPR$
and so:
: $\Box BDNK = \Box KGQR$
So all of them are equal:
: $\Box CBKG = \Box BDNK = \Box KGQR = \Box KNPR$
So the four of them together equal $4 \Box CBKG$.
Similarly, by Parallelograms with Equal Base and Same Height have Equal Area:
: $\Box ACGM = \Box MGQO$
: $\Box QRLH = \Box RPFL$
From Complements of Parallelograms are Equal:
:$\Box MGQO = \Box QRLH$
and:
: $\Box ACGM = \Box RPFL$
So all of them are equal:
: $\Box ACGM = \Box MGQO = \Box QRLH = \Box RPFL$
So the four of them together equal $4 \Box ACGM$.
Adding all these eight areas together, we see that the gnomon $STU$ equals $4 \Box CBKG + 4 \Box ACGM = 4 \Box ABKM$.
But $\Box ABKM$ is the rectangle contained by $AB$ and $BC$, as $BC = BK$.
So four times the rectangle contained by $AB$ and $BC$ equals the area of the gnomon $STU$.
Now we add $\Box QOEH$ to each.
Note that $\Box QOEH$ equals the square on $AC$.
So four times the rectangle contained by $AB$ and $BC$, together with the square on $AC$, equals the gnomon $STU$ together with $\Box QOEH$.
But the gnomon $STU$ together with $\Box QOEH$ form the whole square $ADFE$, which is the square on $AB$ and $BC$ as a single straight line.
Hence the result.
{{qed}}
Theorem: Square of Sum

:$\forall x, y \in \R: \paren {x + y}^2 = x^2 + 2 x y + y^2$

Proof:
Follows from the distribution of multiplication over addition:
{{begin-eqn}}
{{eqn | l = \paren {x + y}^2
      | r = \paren {x + y} \cdot \paren {x + y}
      | c = {{Defof|Square Function}}
}}
{{eqn | r = x \cdot \paren {x + y} + y \cdot \paren {x + y}
      | c = Real Multiplication Distributes over Addition
}}
{{eqn | r = x \cdot x + x \cdot y + y \cdot x + y \cdot y
      | c = Real Multiplication Distributes over Addition
}}
{{eqn | r = x \cdot x + x \cdot y + x \cdot y + y \cdot y
      | c = Real Multiplication is Commutative
}}
{{eqn | r = x \cdot x + 2 x y + y \cdot y
      | c = simplifying
}}
{{eqn | r = x^2 + 2 x y + y^2
      | c = {{Defof|Square Function}}
}}
{{end-eqn}}
{{qed}}
Theorem: Square of Sum

:$\forall x, y \in \R: \paren {x + y}^2 = x^2 + 2 x y + y^2$

Proof:
Follows directly from the Binomial Theorem:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$
putting $n = 2$.
{{qed}}
Theorem: Square of Sum

:$\forall x, y \in \R: \paren {x + y}^2 = x^2 + 2 x y + y^2$

Proof:
{{:Euclid:Proposition/II/4}}
:250px
Let the straight line $AB$ be cut at random at $C$.
Construct the square $ADEB$ on $AB$ and join $DB$.
Construct $CF$ parallel to $AD$ through $C$.
Construct $HK$ parallel to $AB$ through $G$.
From Parallelism implies Equal Alternate Angles, $\angle CGB = \angle ADB$.
We have that $BA = AD$.
So from Isosceles Triangle has Two Equal Angles:
:$\angle ADB = \angle ABD$
So $\angle CGB = \angle CBG$, and so from Triangle with Two Equal Angles is Isosceles, $BC = CG$.
From Opposite Sides and Angles of Parallelogram are Equal, we have $CB = GK$ and $CG = KB$ and so $CGKB$ is equilateral.
Now, since $CG \parallel BK$, from Parallelism implies Supplementary Interior Angles we have that $\angle KBC$ and $\angle GCB$ are supplementary.
But as $\angle KBC$ is a right angle, $\angle BCG$ is also a right angle.
So from Opposite Sides and Angles of Parallelogram are Equal, $\angle CGK$ and $\angle GKB$ are also right angles.
So $CGKB$, described on $CB$, is right-angled.
As $CGKB$ is also equilateral, by definition it is square.
For the same reason, $HGDF$ is also square, and described on $HG$, which equals $AB$.
So $HGDF$ and $CGKB$ are the squares on $AC$ and $CB$.
Now we have from Complements of Parallelograms are Equal that $\Box ACGH = \Box GKEF$.
But $\Box ACGH$ is the rectangle contained by $AC$ and $CB$, as $CB = GC$.
So $\Box GKEF$ is also equal to the rectangle contained by $AC$ and $CB$.
But the squares $HGFD$ and $CBKG$ are equal to the squares on $AC$ and $CB$.
So the four areas $HGFD$, $CBKG$, $ACGH$ and $GKEF$ are equal to:
:the squares on $AC$ and $CB$
and:
:twice the rectangle contained by $AC$ and $CB$.
But $HGFD$, $CBKG$, $ACGH$ and $GKEF$ are also equal to the square on $AB$.
Hence the result.
{{qed}}
Theorem: Square of Sum less Square

:$\forall x, y \in \R: \paren {2x + y} y = \paren {x + y}^2 - x^2$

Proof:
Follows directly and trivially from Square of Sum:
{{begin-eqn}}
{{eqn | l = \paren {x + y}^2
      | r = x^2 + 2 x y + y^2
      | c = Square of Sum
}}
{{eqn | ll= \leadsto
      | l = \paren {x + y}^2 - x^2
      | r = \paren {2 x + y} y
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Square of Sum less Square

:$\forall x, y \in \R: \paren {2x + y} y = \paren {x + y}^2 - x^2$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {x + y}^2 - x^2
      | r = \paren {x + y + x} \paren {x + y - x}
      | c = Difference of Two Squares
}}
{{eqn | r = \paren {2 x + y} y
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Square of Sum less Square

:$\forall x, y \in \R: \paren {2x + y} y = \paren {x + y}^2 - x^2$

Proof:
{{:Euclid:Proposition/II/6}}
Let $AB$ be bisected at $C$, and let $BD$ be added to it in a straight line.
Then the rectangle contained by $AD$ and $BD$ together with the square on $BC$ equals the square on $CD$.
:400px
The proof is as follows.
Construct the square $CEFD$ on $CD$, and join $DE$.
Construct $BG$ parallel to $CE$ through $B$, and let $BG$ cross $DE$ at $H$.
Construct $KM$ parallel to $AD$ through $H$.
Construct $AK$ parallel to $DF$ through $A$.
As $AC = CB$, from Parallelograms with Equal Base and Same Height have Equal Area we have that $\Box ACLK = \Box CBHL$.
From Complements of Parallelograms are Equal, $\Box CBHL = \Box HMFG$.
So $\Box ACLK = \Box HMFG$.
Add $\Box CDML$ to each.
So the whole of $\Box ADMK$ is equal to the gnomon $CDFGHL$.
But $\Box ADMK$ is the rectangle contained by $AD$ and $BD$, because $BD = DM$.
So the gnomon $CDFGHL$ is equal in area to the rectangle contained by $AD$ and $BD$.
Now $\Box LHGE$ is equal to the square on $BC$.
Add $\Box LHGE$ to each of the gnomon $CDFGHL$ and $\Box ADMK$.
Then the gnomon $CDFGHL$ together with $\Box LHGE$ equals the rectangle contained by $AD$ and $BD$ and the square on $BC$.
But the gnomon $CDFGHL$ together with $\Box LHGE$ is the square $CDFE$.
Hence the result.
{{qed}}
Theorem: Square of Difference

:$\forall x, y \in \R: \paren {x - y}^2 = x^2 - 2 x y + y^2$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {x - y}^2 = 
      | r = \paren {x - y} \cdot \paren {x - y}
      | c = 
}}
{{eqn | r = x \cdot \paren {x - y} - y \cdot \paren {x - y}
      | c = Real Multiplication Distributes over Addition
}}
{{eqn | r = x \cdot x - x \cdot y - y \cdot x + y \cdot y
      | c = Real Multiplication Distributes over Addition
}}
{{eqn | r = x^2 - 2xy + y^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Square of Difference

:$\forall x, y \in \R: \paren {x - y}^2 = x^2 - 2 x y + y^2$

Proof:
Follows directly from the Binomial Theorem:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$
putting $n = 2$ and $y = -y$.
{{qed}}
Theorem: Square of Difference

:$\forall x, y \in \R: \paren {x - y}^2 = x^2 - 2 x y + y^2$

Proof:
{{:Euclid:Proposition/II/7}}
:250px
That is: 
:$x^2 + y^2 = \paren {x - y}^2 + 2 x y$
Let the straight line $AB$ be cut at random at $C$.
Construct the square $ADEB$ on $AB$ and join $DB$.
Construct $CN$ parallel to $AD$ through $C$ and let it cross $DB$ at $G$.
Construct $HF$ parallel to $AB$ through $G$.
From Complements of Parallelograms are Equal, $\Box ACGH = \Box FGNE$, so add $\Box CBFG$ to each.
So the whole of $\Box ABFH$ equals the whole of $\Box CBEN$.
So $2 \Box ABFH = \Box ABFH + \Box CBEN$.
But $\Box ABFH + \Box CBEN$ equals the area of the gnomon $KLM$ together with $\Box CBFG$.
So the gnomon $KLM$ together with $\Box CBFG$ equals $2 \Box ABFH$.
But twice the rectangle contained by $AB$ and $BC$ is also equal to $2 \Box ABFH$, as $BG = BC$.
So the gnomon $KLM$ together with $\Box CBFG$ equals twice the rectangle contained by $AB$ and $BC$.
Add $\Box DHGN$ to each.
Note that $\Box DHGN$ equals the square on $AC$.
Then the gnomon $KLM$ together with $\Box CBFG$ and $\Box DHGN$ equals the whole of $\Box ABED$ and $\Box CBFG$, which are the squares on $AB$ and $BC$.
Hence the result.
{{qed}}
Theorem: Square of Sum with Double

:$\forall a, b \in \R: \paren {a + 2 b}^2 = a^2 + 4 a b + 4 b^2$

Proof:
Follows from the distribution of multiplication over addition:
{{begin-eqn}}
{{eqn | l = \paren {a + 2 b}^2
      | r = \paren {a + 2 b} \cdot \paren {a + 2 b}
      | c = 
}}
{{eqn | r = a \cdot \paren {a + 2 b} + 2 b \cdot \paren {a + 2 b}
      | c = Real Multiplication Distributes over Addition
}}
{{eqn | r = a \cdot a + a \cdot 2 b + 2 b \cdot a + 2 b \cdot 2 b
      | c = Real Multiplication Distributes over Addition
}}
{{eqn | r = a^2 + 4 a b + 4 b^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Square of Sum with Double

:$\forall a, b \in \R: \paren {a + 2 b}^2 = a^2 + 4 a b + 4 b^2$

Proof:
A direct application of the Binomial Theorem:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$
putting $n = 2, x = a, y = 2 b$.
{{qed}}
Theorem: Square of Sum with Double

:$\forall a, b \in \R: \paren {a + 2 b}^2 = a^2 + 4 a b + 4 b^2$

Proof:
{{:Euclid:Proposition/II/8}}
That is: $4 \paren {a + b} b + a^2 = \paren {a + 2 b}^2$.
:400px
Let the straight line $AB$ be cut at random at $C$.
Then four times the rectangle contained by $AB$ and $BC$ together with the  square on $AC$ equals the square on $AB$ and $BC$ as a single straight line.
The proof is as follows.
Produce $AB$ to $D$ where $BD = BC$.
Construct the square $AEFD$ on $AD$ and join $DE$.
Draw the given figure above, in the same manner as in Square of Sum.
By Parallelograms with Equal Base and Same Height have Equal Area:
: $\Box CBKG = \Box BDNK$
: $\Box KGQR = \Box KNPR$
From Complements of Parallelograms are Equal:
: $\Box CBKG = \Box KNPR$
and so:
: $\Box BDNK = \Box KGQR$
So all of them are equal:
: $\Box CBKG = \Box BDNK = \Box KGQR = \Box KNPR$
So the four of them together equal $4 \Box CBKG$.
Similarly, by Parallelograms with Equal Base and Same Height have Equal Area:
: $\Box ACGM = \Box MGQO$
: $\Box QRLH = \Box RPFL$
From Complements of Parallelograms are Equal:
:$\Box MGQO = \Box QRLH$
and:
: $\Box ACGM = \Box RPFL$
So all of them are equal:
: $\Box ACGM = \Box MGQO = \Box QRLH = \Box RPFL$
So the four of them together equal $4 \Box ACGM$.
Adding all these eight areas together, we see that the gnomon $STU$ equals $4 \Box CBKG + 4 \Box ACGM = 4 \Box ABKM$.
But $\Box ABKM$ is the rectangle contained by $AB$ and $BC$, as $BC = BK$.
So four times the rectangle contained by $AB$ and $BC$ equals the area of the gnomon $STU$.
Now we add $\Box QOEH$ to each.
Note that $\Box QOEH$ equals the square on $AC$.
So four times the rectangle contained by $AB$ and $BC$, together with the square on $AC$, equals the gnomon $STU$ together with $\Box QOEH$.
But the gnomon $STU$ together with $\Box QOEH$ form the whole square $ADFE$, which is the square on $AB$ and $BC$ as a single straight line.
Hence the result.
{{qed}}
Theorem: Sum of Squares of Sum and Difference

:$\forall a, b \in \R: \paren {a + b}^2 + \paren {a - b}^2 = 2 \paren {a^2 + b^2}$

Proof:
{{begin-eqn}}
{{eqn | o = 
      | r = \left({a + b}\right)^2 + \left({a - b}\right)^2
      | c = 
}}
{{eqn | r = a^2 + 2 a b + b^2 + a^2 - 2 a b + b^2
      | c = Square of Sum and Square of Difference
}}
{{eqn | r = a^2 + b^2 + a^2 + b^2
}}
{{eqn | r = 2 \left({a^2 + b^2}\right)
      | c = 
}}
{{end-eqn}}
{{qed}}