hcju/proofwikiqa huggingface dataset:
Let $\Box AB$ and $\Box BC$ be two equiangular parallelograms of equal area such that the angles at $B$ are equal. Let $DB, BE$ be placed in a straight line. By Two Angles making Two Right Angles make Straight Line it follows that $FB, BG$ also make a straight line. We need to show that $DB : BE = GB : BF$, that is, the sides about the equal angles are reciprocally proportional. :300px Let the parallelogram $\Box FE$ be completed. We have that $\Box AB$ is of equal area with $\Box BC$, and $\Box FE$ is another area. So from Ratios of Equal Magnitudes: : $\Box AB : \Box FE = \Box BC : \Box FE$ But from Areas of Triangles and Parallelograms Proportional to Base: : $\Box AB : \Box FE = DB : BE$ Also from Areas of Triangles and Parallelograms Proportional to Base: : $\Box BC : \Box FE = GB : BF$ So from Equality of Ratios is Transitive: : $DB : BE = GB : BF$ {{qed|lemma}} Next, suppose that $DB : BE = GB : BF$. From Areas of Triangles and Parallelograms Proportional to Base: : $DB : BE = \Box AB : \Box FE$ Also from Areas of Triangles and Parallelograms Proportional to Base: : $GB : BF = \Box BC : \Box FE$ So from Equality of Ratios is Transitive: : $\Box AB : \Box FE = \Box BC : \Box FE$ So from Magnitudes with Same Ratios are Equal: : $\Box AB = \Box BC$ {{qed}} {{Euclid Note|14|VI}}
Let $\triangle ABC, \triangle ADE$ be triangles of equal area which have one angle equal to one angle, namely $\angle BAC = \angle DAE$. We need to show that $CA : AD = EA : AB$, that is, the sides about the equal angles are reciprocally proportional. :250px Place them so $CA$ is in a straight line with $AD$. From Two Angles making Two Right Angles make Straight Line $EA$ is also in a straight line with $AB$. Join $BD$. It follows from Ratios of Equal Magnitudes that: : $\triangle CAB : \triangle BAD = \triangle EAD : \triangle BAD$ But from Areas of Triangles and Parallelograms Proportional to Base: : $\triangle CAB : \triangle BAD = CA : AD$ Also from Areas of Triangles and Parallelograms Proportional to Base: :$\triangle EAD : \triangle BAD = EA : AB$ So from Equality of Ratios is Transitive: : $CA : AD = EA : AB$ {{qed|lemma}} Now let the sides in $\triangle ABC, \triangle ADE$ be reciprocally proportional. That is, $CA : AD = EA : AB$. Join $BD$. From Areas of Triangles and Parallelograms Proportional to Base: : $\triangle CAB : \triangle BAD = CA : AD$ Also from Areas of Triangles and Parallelograms Proportional to Base: : $\triangle EAD : \triangle BAD = EA : AB$ It follows from Equality of Ratios is Transitive that: : $\triangle CAB : \triangle BAD = \triangle EAD : \triangle BAD$ So from Magnitudes with Same Ratios are Equal: : $\triangle ABC = \triangle ADE$ {{qed}} {{Euclid Note|15|VI}}
Let the four straight lines $AB, CD, E, F$ be proportional, that is, $AB : CD = E : F$. What we need to show is that the rectangle contained by $AB$ and $F$ is equal in area to the rectangle contained by $CD$ and $E$. :400px Let $AG, CH$ be drawn perpendicular to $AB$ and $CD$. Let $AG = F$, $CH = E$. Complete the parallelograms $BG$ and $DH$. We have that $AB : CD = E : F$, while $E = CH$ and $F = AG$. So in $\Box BG$ and $\Box DH$ the sides about the equal angles are reciprocally proportional. But from Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional: : $\Box BG = \Box DH$ (in area). We also have that: : $\Box BG$ is the rectangle contained by $AB$ and $F$ : $\Box DH$ is the rectangle contained by $CD$ and $E$ Hence the result. {{qed|lemma}} Now suppose that the rectangle contained by $AB$ and $F$ is equal in area to the rectangle contained by $CD$ and $E$. We use the same construction, and note that $\Box BG = \Box DH$ (in area). But they are equiangular, as all angles are equal to a right angle. So from Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional: : $AB : CD = CH : AG$ But $E = CH$ and $F = AG$. So: : $AB : CD = E : F$ {{qed}} {{Euclid Note|16|VI|It is a special case of {{EuclidPropLink|book=VI|prop=14|title=Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional}}.}}
Let the three straight lines $A, B, C$ be proportional, that is:  :$A : B = B : C$ Then we need to show that the rectangle contained by $A$ and $C$ equals the square on $B$. :300px Let $D = B$. Then $A : B = D : C$ By Rectangles Contained by Proportional Straight Lines, the rectangle contained by $A$ and $C$ equals the rectangle contained by $B$ and $D$. But as $B = D$, the rectangle contained by $B$ and $D$ equals the square on $B$. So the rectangle contained by $A$ and $C$ equals the square on $B$. {{qed|lemma}} Now let the rectangle contained by $A$ and $C$ be equal to the square on $B$. Using the same construction, the rectangle contained by $A$ and $C$ equals the rectangle contained by $B$ and $D$ because $B = D$. So by Rectangles Contained by Proportional Straight Lines: : $A : B = D : C$ But as $B = D$: : $A : B = B : C$ {{qed}} {{Euclid Note|17|VI|This is, of course, just a special case of {{EuclidPropLink|book = VI|prop = 16|title = Rectangles Contained by Proportional Straight Lines}}.}}
From Sum of Angles of Triangle Equals Two Right Angles $\angle CFD = \angle AGB$. So $\triangle FCD$ is equiangular with $\triangle GAB$. So from Equiangular Triangles are Similar, $\triangle FCD$ is similar to $\triangle GAB$. So $FD : GB = FC : GA = CD : AB$. Similarly from Sum of Angles of Triangle Equals Two Right Angles $\angle GHB = \angle FED$. So $\triangle FDE$ is equiangular with $\triangle GBH$. So from Equiangular Triangles are Similar, $\triangle FDE$ is similar to $\triangle GBH$. So $FD : GB = FE : GH = ED : HB$. Thus we have that: :$FC : AG = CD : AB = FE : GH = ED : HB$ Since $\angle CFD = \angle AGB$ and $\angle DFE = \angle BGH$, we have that :$\angle CFE = \angle CFD + \angle DFE = \angle AGB + \angle BGH = \angle AGH$ For the same reason: :$\angle CDE = \angle CDF + \angle FDE = \angle ABG + \angle GBH = \angle ABH$ So $CDEF$ is equiangular with $ABHG$. As has been shown, the sides of these polygons are proportional about their equal angles. So from {{EuclidDefLink|VI|1|Similar Rectilineal Figures}}, $CDEF$ is similar $ABHG$. {{qed}} {{Euclid Note|18|VI|In {{EuclidPropLink|book=VI|prop=20|title=Similar Polygons are composed of Similar Triangles}}, it is shown by dividing any polygon into triangles, any two similar polygons are composed of similar triangles.<br/>Thus the construction as given here can be seen directly to extend to polygons with any number of sides.}}
Let $\triangle ABC$ and $\triangle DEF$ be triangles such that $\angle ABC = \angle DEF$ and $\angle BAC = \angle EDF$. Then from Sum of Angles of Triangle Equals Two Right Angles $\angle ACB$ is equal to two right angles minus $\angle ABC + \angle BAC$. Also from Sum of Angles of Triangle Equals Two Right Angles $\angle DFE$ is equal to two right angles minus $\angle DEF + \angle EDF$. That is, $\angle DFE$ is equal to two right angles minus $\angle ABC + \angle BAC$. So $\angle DFE = \angle ACB$ and so all three corresponding angles of $\triangle ABC$ and $\triangle DEF$ are equal. The result follows from Equiangular Triangles are Similar. {{qed}}
Let $\triangle ABC$ and $\triangle DEF$ be similar, such that $\angle ABC = \angle DEF$ and $AB : BC = DE : EF$ such that $BC$ corresponds to $EF$. :400px Let $BG$ be constructed such that $EF : BG = BC : EF$, and join $AG$. From Proportional Magnitudes are Proportional Alternately $AB : DE = BC : EF$. So from Equality of Ratios is Transitive $AB : DE = EF : BG$. So in $\triangle ABC$ and $\triangle DEF$ the sides about the equal angles are reciprocally proportional. From Sides of Equiangular Triangles are Reciprocally Proportional, the area of $\triangle ABG$ equals the area of $\triangle DEF$. Now we have that $BC : EF = EF : BG$. So from {{EuclidDefLink|V|9|Duplicate Ratio}} $BC$ has to $BG$ a ratio duplicate to that which $CB$ has to $EF$. But from Areas of Triangles and Parallelograms Proportional to Base, $CB : BG = \triangle ABC : \triangle ABG$. So $\triangle ABC$ has to $\triangle ABG$ a ratio duplicate to that which $BC$ has to $EF$. But $\triangle ABC = \triangle DEF$. So $\triangle ABC$ has to $\triangle DEF$ a ratio duplicate to that which $BC$ has to $EF$. {{qed}}
Let $ABCDE$ and $FGHKL$ be similar polygons such that $AB$ corresponds to $FG$. We need to show that $ABCDE$ and $FGHKL$ are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes. Also that the area of the polygon $ABCDE$ has to the polygon $FGHKL$ a ratio duplicate of $AB : FG$. :450px Join up $BE, EC, GL, LH$. Since $ABCDE$ and $FGHKL$ are similar: : $\angle BAE = \angle GFL$ From {{EuclidDefLink|VI|1|Similar Rectilineal Figures}}: : $BA : AE = GF : FL$ Thus from Triangles with One Equal Angle and Two Sides Proportional are Similar, $\triangle ABE$ is similar to $\triangle FGL$. So $\angle ABE = \angle FGL$. But $\angle ABC = \angle FGH$ because $ABCDE$ and $FGHKL$ are similar. So: : $\angle EBC = \angle LGH$ Because $\triangle ABE$ is similar to $\triangle FGL$: : $EB : BA = LG : GF$ Also, because $ABCDE$ and $FGHKL$ are similar: : $AB : BC = FG : GH$ So from Equality of Ratios Ex Aequali: : $EB : BC = LG : GH$ So from Triangles with One Equal Angle and Two Sides Proportional are Similar, $\triangle EBC$ is similar to $\triangle LGH$. For the same reason, $\triangle ECD$ is similar to $\triangle LHK$. So $ABCDE$ and $FGHKL$ have been divided into similar triangles, and into triangles equal in multitude. Now let $AC, FH$ be joined. Because $ABCDE$ and $FGHKL$ are similar: : $\angle ABC = \angle FGH$ From Triangles with One Equal Angle and Two Sides Proportional are Similar $\triangle ABC$ is similar to $\triangle FGH$. Therefore $\angle BAC = \angle GFH$ and $\angle BCA = \angle GHF$. Also, we have that $\angle BAM = \angle GFN$, and $\angle ABM = \angle FGN$. So from Sum of Angles of Triangle Equals Two Right Angles $\angle AMB = \angle FGN$ and so $\triangle ABM$ is similar to $\triangle FGN$. Similarly we can show that $\triangle BMC$ is similar to $\triangle GNH$. Therefore $AM : MB = FN : NG$ and $BM : MC = FN : NH$. But from Areas of Triangles and Parallelograms Proportional to Base: : $AM : MC = \triangle ABM : \triangle MBC$. So from Sum of Components of Equal Ratios: : $\triangle ABM : \triangle MBC = \triangle ABE : \triangle CBE$ But: : $\triangle ABM : \triangle MBC = AM : MC$ So: : $\triangle ABE : \triangle CBE = AM : MC$ For the same reason: : $FN : NH = \triangle FGL : \triangle GLH$ As $AM : MC = FN : NH$, it follows that: : $\triangle ABE : \triangle BEC = \triangle FGL : \triangle GLH$ Similarly: : $\triangle ABE : \triangle FGL = \triangle BEC : \triangle GLH$ We now join $BD$ and $GK$, and by a similar construction show that: : $\triangle BEC : \triangle LGH = \triangle ECD : \triangle LHK$ From Sum of Components of Equal Ratios: : $ABE : FGL = ABCDE : FGHKL$ But from Ratio of Areas of Similar Triangles $\triangle ABE$ has to $\triangle FGL$ a ratio duplicate of $AB : FG$. Therefore the area of the polygon $ABCDE$ has to the polygon $FGHKL$ a ratio duplicate of $AB : FG$. {{qed}} === Porism === {{:Similar Polygons are composed of Similar Triangles/Porism}} {{Euclid Note|20|VI}}
:500px We have that $A$ is similar to $C$. From {{EuclidDefLink|VI|1|Similar Rectilineal Figures}}, it is equiangular with it and the sides about the equal angles are proportional. We also have that $B$ is similar to $C$. Again, from {{EuclidDefLink|VI|1|Similar Rectilineal Figures}}, it is equiangular with it and the sides about the equal angles are proportional. So by definition $A$ is similar to $B$. The statements of reflexivity and symmetry are shown similarly. It follows that if $A$ is similar to $B$, and $B$ is similar to $C$, then $A$ is similar to $C$. Thus similarity between polygons is transitive. Hence the result, by definition of equivalence relation. {{qed}} {{Euclid Note|21|VI|{{AuthorRef|Euclid}} himself did not have the concept of an equivalence relation.<br/>However, the extra statements leading to the main result are sufficiently straightforward to justify adding the full proof here.}}
Let the four straight lines $AB, CD, EF, GH$ be proportional. That is: : $AB : CD = EF : GH$ Let there be described on $AB$ and $CD$ the similar and similarly situated rectilineal figures $KAB$ and $LCD$. Let there be described on $EF$ and $GH$ the similar and similarly situated rectilineal figures $MF$ and $NH$. We need to show that: : $KAB : LCD = MF : NH$ By Construction of Third Proportional Straight Line, let there be taken a third proportional $O$ to $AB, CD$, and a third proportional $P$ to $EF, GH$. We have that: : $AB : CD = EF : GH$, and $CD : O = GH : P$ So from Equality of Ratios Ex Aequali: : $AB : O = EF : P$ But from {{EuclidPropLink|book=VI|prop=19|qual=Porism|title=Ratio of Areas of Similar Triangles}}: : $AB : O = KAP : LCD$ and: : $EF : P = MF : NH$ Therefore from Equality of Ratios is Transitive: : $KAB : LCD = MF : NH$ :500px Now suppose $MF : NH = KAB : LCD$. We need to show that: : $AB : CD = EF : GH$ Suppose, with a view to obtaining a contradiction, that: : $EF : GH \ne AB : CD$ Instead, by Construction of Fourth Proportional Straight Line, let: : $EF : QR = AB : CD$ On $QR$, by Construction of Similar Polygon, let the rectilineal figures $SR$ be described similar and similarly situated to either $MF$ or $NH$. We have that: : $AB : CD = EF : QR$ We also have that there have been described on $AB$ and $CD$ the similar and similarly situated rectilineal figures $KAB$ and $LCD$ We also have that there have been on $EF$ and $GH$ the similar and similarly situated rectilineal figures $MF$ and $NH$. From Equality of Ratios is Transitive: : $MF : SF = MF : NH$ Therefore $MF$ has the same ratio to each of the figures $NH, SR$. Therefore, by Magnitudes with Same Ratios are Equal: : $NH = SR$ But it is also similar and similarly situated to it. Therefore: : $GH = QR$ But we have that $AB : CD = EF : QR$ while $QR = GH$. Therefore: : $AB : CD = EF : GH$ {{Qed}} {{Euclid Note|22|VI}}
Let $\Box AC, \Box CF$ be equiangular parallelograms having $\angle BCD = \angle ECG$. Let them be arranged so that $BC$ is in a straight line with $CG$. Then $DC$ is in a straight line with $CE$. :400px Let the parallelogram $DG$ be completed. Let a straight line $K$ be set out. Using Construction of Fourth Proportional Straight Line, let $L$ and $M$ be constructed such that: :$K : L = BC : CG$ :$L : M = DC : CE$ Then $K : L$ and $L : M$ are the same as the ratios of the sides. But $K : M$ is compounded of $K : L$ and $L : M$. So $K : M$ is the ratio compounded of the ratios of the sides. From Areas of Triangles and Parallelograms Proportional to Base: : $BC : CG = \Box AC : \Box CH$ From Equality of Ratios is Transitive: : $K : L = \Box AC : \Box CH$ Similarly, from Areas of Triangles and Parallelograms Proportional to Base: : $DC : CE = \Box CH : \Box CF$ Since we have $DC : CE = L : M$, it follows from Equality of Ratios is Transitive that: : $L : M = \Box CH : \Box CF$ Since we have: :$K : L = \Box AC : \Box CH$ :$L : M = \Box CH : \Box CF$ it follows that: :$K : M = \Box AC : \Box CF$ But $K : M$ is the ratio compounded of the ratios of the sides. Hence the result. {{qed}} {{Euclid Note|23|VI}}
Let $\Box ABCD$ be a parallelogram and $AC$ one of its diameters. Let $EG, HK$ be parallelograms about $AC$. :300px We have that $EF \parallel CD$. So from Parallel Line in Triangle Cuts Sides Proportionally: : $BE : EA = CF : FA$ Again, we have that: : $FG \parallel CD$ So from Parallel Line in Triangle Cuts Sides Proportionally: : $CF : FA = DG : GA$ From Equality of Ratios is Transitive: : $BE : EA = DG : GA$ So from Magnitudes Proportional Separated are Proportional Compounded: : $BA : AE = DA : AG$ From Proportional Magnitudes are Proportional Alternately: : $BA : AD = EA : AG$ Therefore in the parallelograms $\Box ABCD$ and $\Box EG$, the sides about the common angle $\angle BAD$ are proportional. We have that $GF \parallel DC$, $\angle AFG = \angle DCA$ and $\angle DAC$ is common to $\triangle ADC$ and $\triangle AGF$. Therefore $\triangle ADC$ is equiangular with $\triangle AGF$. For the same reason, $\triangle ACB$ is equiangular with $\triangle AFE$. Thus the whole parallelogram $\Box ABCD$ is equiangular with the parallelogram $\Box EG$. Therefore: :$AD : DC = AG : GF$ :$DC : CA = GF : FA$ :$AC : CB = AF : FE$ :$CB : BA = FE : FA$ Since we also have: :$DC : CA = GF : FA$ :$AC : CB = AF : FE$ it follows from Equality of Ratios Ex Aequali that: : $DC : CB = GF: FE$ Therefore in the parallelograms $\Box ABCD$ and $\Box EG$, sides about the equal angles are proportional. Therefore from {{EuclidDefLink|VI|1|Similar Rectilineal Figures}}, $\Box ABCD$ is similar to $\Box EG$. By the same argument, $\Box ABCD$ is similar to $\Box KH$. Therefore by Similarity of Polygons is Equivalence Relation, $\Box EG$ is similar to $\Box HK$. {{qed}} {{Euclid Note|24|VI|{{EuclidNoteConverse|prop=26|title=Parallelogram Similar and in Same Angle has Same Diameter}}}}
We have that $BC$ is in a straight line with $CF$ and that $LE$ is in a straight line with $EM$. From the porism to Ratio of Areas of Similar Triangles: : $BC : CF = \triangle ABC : \triangle KGH$ But from Areas of Triangles and Parallelograms Proportional to Base: : $BC : CF = \Box BE : \Box EF$ So: : $\triangle ABC : \triangle KGH = \Box BE : \Box EF$ So from Proportional Magnitudes are Proportional Alternately: : $\triangle ABC : \Box BE = \triangle KGH : \Box EF$ But: : $\triangle ABC = \Box BE$ and so: : $\triangle KGH = \Box EF$ But $\Box EF = D$, and so $\triangle KGH = D$ Also we have that $\triangle KGH$ is similar to $\triangle ABC$. Hence the result. {{qed}} {{Euclid Note|25|VI}}
From the parallelogram $\Box ABCD$ let there be taken away the parallelogram $\Box AF$ similar and similarly situated to the whole and having a common angle with it. We need to show that $\Box ABCD$ is about the same diameter with $\Box AF$. :400px Suppose it is not, and that $AHC$ is the diameter of $\Box ABCD$. Let $GF$ be produced to $H$. From Construction of Parallel Line let $HK$ be drawn parallel to $AD$ and $BC$. From Parallelograms About Diameter are Similar: : $DA : AB = GA : AK$ But also because of the similarity of $\Box ABCD$ and $\Box EG$: : $DA : AB = GA : AE$ So from Equality of Ratios is Transitive: : $GA : AK = GA : AE$ Therefore from Magnitudes with Same Ratios are Equal: : $AK = AE$ Hence the result. {{Qed}} {{Euclid Note|26|VI|{{EuclidNoteConverse|prop=24|title=Parallelograms About Diameter are Similar}}}}
Let $AB$ be a straight line and let it be bisected at $C$. Let the parallelogram $\Box AD$ be applied on $AB$ deficient by the parallelogramic figure $\Box DB$ described on $AC$. We need to show that of all the parallelograms applied to $AB$ and deficient by parallelogramic figures similar and similarly situated to $\Box DB$, $\Box AD$ is the greatest. :400px Let $\Box AF$ be applied to $AB$ deficient by $\Box FB$ similar and similarly situated to $\Box DB$. We have that $\Box DB$ is similar to $\Box FB$. So from Parallelogram Similar and in Same Angle has Same Diameter they are about the same diameter. Let that diameter $DB$ be drawn, and let that figure be described. From Complements of Parallelograms are Equal: : $\Box CF = \Box FE$ As $\Box FB$ is common: : $\Box CH = \Box KE$ But from Parallelograms with Equal Base and Same Height have Equal Area: : $\Box CH = \Box CG$ since $AC = CB$ Therefore: : $\Box GC = \Box EK$ Add $\Box CF$ to each. Therefore $\Box AF$ is equal in area to the gnomon $LMN$. So $\Box DB$, that is $\Box AD$, is greater than $\Box AF$. Hence the result. {{qed}} {{Euclid Note|27|VI}}
Let $AG = C$. Then $\Box AG$ has been applied to $AB$ equal in area to $C$ and deficient by $\Box GB$ similar to $D$. Otherwise let $\Box HE > C$. Now $HE = GB$, so: : $GB > C$ We have that $\Box KLMN$ is constructed equal to the excess by which $GB$ is greater than $C$ and similar and similarly situated to $D$. But $D$ is similar to $GB$ and so from Similarity of Polygons is Equivalence Relation $KM$ is also similar to $GB$. Let $KL$ correspond to $GE$ and $LM$ to $GF$. Since $\Box GB = C + \Box KLMN$ it follows that: : $\Box GB > \Box KLMN$ Therefore $GE > KL$ and $GF > KM$. We have $GO$ constructed equal to $KL$ and $GP$ equal to $KM$. Therefore $\Box OGPQ = \Box KLMN$ and $\Box OGPQ$ is similar to $\Box KLMN$. Therefore from Similarity of Polygons is Equivalence Relation $\Box GQ$ is similar to $\Box GB$. So from Parallelogram Similar and in Same Angle has Same Diameter, $GQ$ is about the same diameter with $GB$. Let $GQB$ be this diameter and let the figure be described. Since $\Box BG = C + \Box KM$, and in them $\Box GQ = \Box KM$, it follows that the gnomon $UWV$ equals the remainder $C$. From [[Complements of Parallelograms are Equal]: : $\Box PR = \Box OS$. Let $\Box QB$ be added to each. Then: : $\Box PB = \Box OB$ But since $AE = EB$, it follows from Parallelograms with Equal Base and Same Height have Equal Area that: : $\Box OB = \Box TE$ Therefore: : $\Box TE = \Box PB$ Let $\Box OS$ be added to each. Therefore the whole of $\Box TS$ equals the area of the gnomon $UWV$. But $UWV$ was shown to be equal to $C$. Hence the result. {{qed}} {{Euclid Note|28|VI}}
Let $KH$ correspond to $FL$ and $KG$ to $FE$. Since $GH > FB$, we have that $KH > FL$ and $KG > FE$. Since $FLM = KH$ and $FEN = KG$, it follows that $\Box MN = \Box GH$ and also $\Box MN$ is similar to $\Box GH$. But $\Box GH$ is similar to $\Box EL$. So from Similarity of Polygons is Equivalence Relation, $\Box MN$ is similar to $\Box EL$. Therefore from Parallelogram Similar and in Same Angle has Same Diameter $\Box EL$ is about the same diameter with $\Box MN$. Let that diameter $FO$ be drawn, and the figure described. Since $\Box GH = \Box EL + C$, while $\Box MN = \Box GH$, it follows that $\Box MN = \Box EL + C$. Subtract $\Box EL$ from each. Then the gnomon $XWV$ is equal to $C$. Since $AE = EB$, from Parallelograms with Equal Base and Same Height have Equal Area, $\Box AN = \Box NB$. Thus, from Complements of Parallelograms are Equal, $\Box AN = \Box LP$. Add $\Box EO$ to each. Then $\Box AO$ equals the gnomon $XWV$. But the gnomon $XWV$ equals $C$. So $\Box AO = C$. Hence the result. {{qed}} {{Euclid Note|29|VI}}
As $\Box BC$ is a square, then $\Box AD$ is also a square. Since $\Box CD = \Box BC$, we subtract $\Box CE$ from each. Therefore $\Box BF = \Box AD$. From Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional, in $\Box BF$ and $\Box AD$ the sides about the equal angles are reciprocally proportional. Therefore $FE : ED = AE : EB$. But $FE = AB$ and $ED = AE$. So $BA : AE = AE : EB$. Also $AB > AE$ and so $AE > EB$. Hence the result. {{qed}} {{Euclid Note|30|VI}}
Let $\triangle ABC$ be a right-angled triangle where $\angle BAC$ is a right angle. We need to show that the area of the figure on $BC$ is equal to the sum of the areas of the similar figures on $AB$ and $AC$. :300px Let $AD$ be drawn perpendicular to $BC$. From Perpendicular in Right-Angled Triangle makes two Similar Triangles, $\triangle ABD$ and $\triangle ADC$ are similar both to each other and to $\triangle ABC$. So from {{EuclidDefLink|VI|1|Similar Rectilineal Figures}}, $CB : BA = AB : BD$. So from the porism to Ratio of Areas of Similar Triangles, the figures on each of these sides are proportional in the same way. So $BC : BD$ is the same as the ratio of the areas of the figures on $CB$ and $BA$. Similarly, $BC : CD$ is the same as the ratio of the areas of the figures on $CB$ and $CA$. So $BC : BD + DC$ equals the ratio of the areas of $BC$ and the sum of the areas of the figures on $CA$ and $AB$. {{qed}} {{Euclid Note|31|VI|Also see {{EuclidPropLink|book=I|prop=47|title=Pythagoras's Theorem}}, of which this is an extension.}}