hcju/proofwikiqa huggingface dataset:
: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 Equality 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}}
: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 Equality: : $\triangle ABC = \triangle DCB$ 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 Common Notion 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 Equality: : $\triangle ABC = \triangle BCD$ So $BC$ bisects the parallelogram. Similarly, $AD$ also bisects the parallelogram. {{qed}} {{Euclid Note|34|I|The use of Triangle Side-Angle-Side Equality in this proof seems to be superfluous as the triangles were already shown to be equal using Triangle Angle-Side-Angle Equality.  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 Equality 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 Equality is used to prove the other congruence theorems, this conclusion would seem to be follow trivially in those cases.}}
:400px Let $ABCD$ and $EBCF$ be parallelograms on the same base and in the same parallels $AF$ and $BC$. As $ABCD$ is a parallelogram then $AD = BC$ from Opposite Sides and Angles of Parallelogram are Equal. For the same reason, $EF = BC$. So by Common Notion 1 we have that $AD = EF$. Also, $DE$ is common, so the whole of $AE$ equals the whole of $DF$, by Common Notion 2. But from Opposite Sides and Angles of Parallelogram are Equal, we have $AB = DC$. From Parallelism implies Equal Corresponding Angles: :$\angle FDC = \angle EAB$ So by Triangle Side-Angle-Side Equality: :$EB = FC$ and so $\triangle EAB = \triangle FDC$. Subtract $\triangle DGE$ from each. Then, by Common Notion 3, the trapezoid $ABGD$ equals the trapezoid $EGCF$. Now we add $\triangle GBC$ to both. Then, by Common Notion 2, $ABCD = EBCF$. {{qed}} {{Euclid Note|35|I}}
:400px Let $ABCD$ and $EFGH$ be parallelograms with equal bases $BC$ and $FG$, and in the same parallels $AH$ and $BG$. Join $BE$ and $CH$. We have $BC = FG$ and $FG = EH$ So by Common Notion 1 we have $BC = EH$. But $EB$ and $HC$ join them. So by Lines Joining Equal and Parallel Straight Lines are Parallel, $EB$ and $HC$ are equal and parallel. So $EBCH$ is a parallelogram. So, by Parallelograms with Same Base and Same Height have Equal Area, the area of $EBCH$ equals the area of $ABCD$. Also by Parallelograms with Same Base and Same Height have Equal Area, the area of $EBCH$ equals the area of $EFGH$. So by Common Notion 1, the area of $ABCD$ equals the area of $EFGH$. {{qed}} {{Euclid Note|36|I}}
:400px Let $ABC$ and $DBC$ be triangles which are on the same base $BC$ and in the same parallels $AD$ and $BC$. Let $AD$ be produced in the directions of $E$ and $F$. Let $BE$ through $B$ be drawn parallel to $CA$. Let $CF$ through $C$ be drawn parallel to $BD$. Then each of $EBCA$ and $DBCF$ 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 $EBCA$ as $AB$ bisects it. For the same reason, $DBC$ is half of $DBCF$. But by Common Notion 1, $\triangle ABC = \triangle DBC$. {{qed}} {{Euclid Note|37|I}}
:400px 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, $\triangle ABC = \triangle DEF$. {{qed}} {{Euclid Note|38|I}}
: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}}.}}
: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.}}
: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}}
: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}}
:400px Let $ABCD$ be a parallelogram, and let $AC$ be a diameter. Let $EKHA$ and $FKGC$ be parallelograms about $AC$. Let $BEKG$ and $DFKH$ be the complements of $EKHA$ and $FKGC$. From Opposite Sides and Angles of Parallelogram are Equal: : $\triangle ABC = \triangle ACD$ Also from Opposite Sides and Angles of Parallelogram are Equal: : $\triangle AEK = \triangle AHK$ : $\triangle KFC = \triangle KGC$ So from Common Notion 2: : $\triangle AEK + \triangle KGC = \triangle AHK + \triangle KFC$ But the whole of $\triangle ABC$ equals the whole of $\triangle ACD$. So when $\triangle AEG + \triangle GIC$ and $\triangle AHG + \triangle GFC$ are subtracted from $\triangle ABC$ and $\triangle ACD$ respectively, the complement $BEKG$ which remains is equal in area to $DFKH$, from Common Notion 3. {{qed}} {{Euclid Note|43|I}}
:600px Let $AB$ be the given line segment, $C$ be the given triangle and let $D$ be the given angle. From Construction of Parallelogram equal to Triangle in Given Angle, construct the parallelogram $BEFG$ equal to $C$ such that $\angle EBG = \angle D$. Place it so that $BE$ is in a straight line with $AB$. Let $FG$ be produced to $H$, and draw $AH$ parallel to $BG$. Then join $HB$. We have that $HF$ falls on the parallel lines $AH$ and $EF$. So from Parallelism implies Supplementary Interior Angles, $\angle AHF$ and $\angle HFE$ are supplementary. So $\angle BHG + \angle GFE$ is less than two right angles. From the Fifth Postulate, $HB$ and $FE$ will meet if produced. So let them meet at $K$. Draw $KL$ parallel to $EA$, and produce $HA$ and $GB$ to the points $L$ and $M$. Then $HLKF$ is a parallelogram. From Complements of Parallelograms are Equal, $ABML$ has the same area as $BEFG$. But $BEFG$ has the same area as $\triangle C$. So by common notion 1, so does $ABML$. From Two Straight Lines make Equal Opposite Angles, $\angle GBE = \angle ABM$, while $\angle GBE = \angle D$. Therefore $ABML$ is the required parallelogram. {{qed}} {{Euclid Note|44|I|There is a suggestion that this result is the work of {{AuthorRef|Pythagoras of Samos|Pythagoras}}.}}
: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 common notion 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 common notion 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.}}
:200px Let $AB$ be the given line segment. Construct $AC$ perpendicular to $AB$. By Construction of Equal Straight Lines from Unequal, place $D$ on $AC$ so $AD = AB$. Construct $DE$ parallel to $AB$, and construct $BE$ parallel to $AD$. So $ADEB$ is a parallelogram. So $AB = DE$ and $AD = BE$ from Opposite Sides and Angles of Parallelogram are Equal. So the parallelogram is equilateral. Also, we have that $AD$ falls on the parallels $AB$ and $DE$. So from Parallelism implies Supplementary Interior Angles, we have that $\angle BAD + \angle ADE$ equals two right angles. But as $\angle BAD$ is right, so is $\angle ADE$. And, from Opposite Sides and Angles of Parallelogram are Equal, so are $\angle ABE$ and $\angle BED$. So $ADEB$ is equilateral and equiangular, and therefore, by definition, a square. {{qed}} {{Euclid Note|46|I}}
:250px Let $AB$ be the given straight line cut at random at the point $C$. Construct the square $ABED$ on $AB$. Construct $CF$ parallel to $AD$. Then $\Box ABED = \Box ACFD + \Box CBEF$. Now $\Box ABED$ is the square on $AB$. Similarly, from Opposite Sides and Angles of Parallelogram are Equal: : $\Box ACDF$ is the rectangle contained by $AB$ and $AC$, as $AB = AD$ : $\Box CBFE$ is the rectangle contained by $AB$ and $BC$, as $AB = AD$. Hence the result. {{qed}} {{Euclid Note|2|II|This is little more than an example of {{EuclidPropLink|prop=1|title=Real Multiplication Distributes over Addition}}, and could be directly derived from it. Euclid, for some reason, preferred not to do this.}}
:250px Let $AB$ be the given straight line cut at random at the point $C$. Construct the square $CDEB$ on $AB$. Produce $ED$ to $F$ and construct $AF$ parallel to $CD$. Then $\Box ABEF = \Box ACDF + \Box CBED$. Now $\Box ABEF$ is the rectangle contained by $AB$ and $BC$, as $BC = CD = AF$. Similarly, from Opposite Sides and Angles of Parallelogram are Equal: : $\Box ACDF$ is the rectangle contained by $AC$ and $BC$, as $BC = AF$ : $\Box CBED$ is the square on $BC$. So the rectangle contained by $AB$ and $BC$ equals the rectangle contained by $AC$ and $BC$ together with the square on $BC$. {{qed}} {{Euclid Note|3|II|This is little more than an example of {{EuclidPropLink|prop=1|title=Real Multiplication Distributes over Addition}}, and could be directly derived from it. Euclid, for some reason, preferred not to do this.}}
:250px Let $\triangle ABC$ be a triangle such that the square on $BC$ equals the sum of the squares on $AB$ and $AC$. Construct $AD$ perpendicular to $AC$. Make $AD$ equal to $AB$, and join $CD$. Since $DA = AB$, the square on $AD$ equals the square on $AB$. Add the square on $AC$ to each. Then the squares on $AD$ and $AC$ equal the squares on $AB$ and $AC$. But $\angle DAC$ is a right angle. Therefore, by Pythagoras's Theorem, the square on $DC$ equals the sum of the squares on $AD$ and $AC$. Also, the  square on $BC$ equals the sum of the squares on $AB$ and $AC$, by hypothesis. So the square on $DC$ equals the square on $BC$, and so $DC = BC$. So from Triangle Side-Side-Side Equality, $\triangle ABC = \triangle DAC$. But as $\triangle DAC$ is a right triangle, then so must $\triangle ABC$ be. {{qed}} {{Euclid Note|48|I|{{EuclidNoteConverse|prop = 47|title = Pythagoras's Theorem}}.}}
Follows from the distribution of multiplication over addition: {{begin-eqn}} {{eqn | l = \left({x + y}\right)^2       | r = \left({x + y}\right) \cdot \left({x + y}\right) }} {{eqn | r = x \cdot \left({x + y}\right) + y \cdot \left({x + y}\right)       | 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}}
{{: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, and as it is 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}}
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}}
{{: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}}
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}}
A direct application of the Binomial Theorem: :$\displaystyle \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$ putting $n = 2, x = a, y = 2 b$. {{qed}}
{{: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}}
Follows directly from the Binomial Theorem: :$\displaystyle \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$ putting $n = 2$. {{qed}}
{{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}}
{{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}}
{{: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 $2 \Box ABGH$. 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}}
Follows directly from the Binomial Theorem: :$\displaystyle \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$ putting $n = 2$ and $y = -y$. {{qed}}