hcju/proofwikiqa huggingface dataset:
{{Questionable|Not only the proof is faulty, this theorem is wrong. If we have a recursively enumerable set A of axioms, then the set of theorems (and hence, the Gödel numbers of those) proven by A is recursively enumerable. Minimal arithmetic and PA are both recursively enumerable, and hence their theorems have a recursively enumerable set of Gödel numbers. Any recursively enumerable set is arithmetical with degree Sigma^0_1 in the arithmetical hierarchy. Moreover, there is already a well known provability predicate Pr(x) in the language of Peano arithmetic. Indeed, if we use the diagonal lemma on ~Pr(x) to get a sentence with T proves G iff ~Pr(G), then we will have an unprovable sentence.|Burak}} The proof is by contradiction. Let $\Theta$ be the set of Gödel numbers of the theorems of $T$. Suppose it is defined in $T$ by the formula $\map \theta y$. Since $T$ contains $Q$, we may apply the Diagonal Lemma to $\neg \map \theta y$. This gives us a sentence $G$ such that :$T \vdash G \leftrightarrow \neg \map \theta {\hat G}$ where $\hat G$ is the Gödel number of $G$ (more accurately, it is the term in the language of arithmetic obtained by applying the function symbol $s$ to $0$ this many times). {{Questionable|The below argument is not correct. $T$ does not necessarily prove that $\neg \map \theta {\hat G}$ just because it does not prove $G$. PA has a provability predicate $\map \Pr x$ which satisfies the property that if PA proves G, then PA proves $\map \Pr {\hat G}$). The step used here assumes a somewhat similar property in the opposite direction that our provability predicate satisfies if T does not prove G, then T proves $\neg \map \theta {\hat G}$, which we did not have as an assumption on $\map \Theta x$. |Burak}} Suppose $G$ is not a theorem of $T$. :Then the Gödel number of $G$ is not in $\Theta$. :Since $\theta$ defines $\Theta$ in $T$, this means that: ::$T \vdash \neg \map \theta {\hat G}$ :But, by choice of $G$ (specifically, the bi-implication above), this gives us: ::$T\vdash G$ :which contradicts $G$ not being a theorem of $T$ Thus, $G$ is a theorem of $T$. :But, then the Gödel number of $G$ is in $\Theta$, and :since $\theta$ defines $\Theta$ in $T$, this means that ::$T \vdash \map \theta {\hat G}$ :But, then this gives us ::$T \vdash \neg G$ :which contradicts $G$ being a theorem of $T$, ''since $T$ is consistent''. Since assuming $\Theta$ was definable in $T$ necessarily leads to a contradiction, we conclude that it is impossible. {{qed}}
There is a primitive recursive function $\mathrm {diag}$ which is defined by: :$\map {\mathrm {diag} } n = \widehat {\map A {\hat A} }$ where: :$\map A x$ is the formula such that $\hat A = n$. :the $\hat{ }$ sign denotes the Gödel number of the contained formula (and we are not being formal about distinguishing between integers and symbols in the language). Informally, $\mathrm{diag}$ takes a Gödel number, decodes it to a formula, plugs in the Gödel number for that formula in place of a free variable, and encodes this new formula back to a new Gödel number. Since $T$ contains $Q$, by Recursive Sets are Definable in Arithmetic applied to the graph of $\mathrm {diag}$, we have that there is some formula $\map {\mathrm {Diag} } {x, y}$ which defines the graph of $\mathrm {diag}$ in $T$. That is: :$\map {\mathrm {diag} } n = m$ if and only if $T\vdash \map {\mathrm {Diag} } {n, m}$ Let $\map A x$ be the formula: :$\exists y \paren {\map {\mathrm {Diag} } {x, y} \land \map B y}$ Let $G$ be $\map A {\hat A}$. We then have $T \vdash \map {\mathrm {Diag} } {\hat A, \hat G}$, by checking the definitions. Let $T' = T \cup \set G$. Then: :$T' \vdash \map A {\hat A}$ Hence: :$T' \vdash \exists y \paren {\map {\mathrm {Diag} } {\hat A, y} \land \map B y}$ But since $\hat G$ is the only number such that $T \vdash \map {\mathrm {Diag} } {\hat A, \hat G}$, this gives us: :$T' \vdash \map B {\hat G}$ Thus: :$T \vdash G \rightarrow \map B {\hat G}$ Let $T' = T \cup \set {\map B {\hat G} }$. Again, we have: :$T \vdash \map {\mathrm {Diag} } {\hat A, \hat G}$ so this gives us: :$T' \vdash \map {\mathrm {Diag} } {\hat A, \hat G} \land \map B {\hat G}$ and hence: :$T' \vdash \exists y \paren {\map {\mathrm {Diag} } {\hat A, y} \land \map B y}$ But this is the same thing as: :$T' \vdash G$ Thus: :$T \vdash \map B {\hat G} \rightarrow G$ Thus $G$ is as claimed. {{qed}} {{MissingLinks|first line of proof}}
$\operatorname {Th}_\ZZ$ is easily seen to be a consistent extension of minimal arithmetic.  (In fact, the axioms in minimal arithmetic were selected based on the behavior of standard arithmetic.) Thus, the theorem is a special case of Set of Gödel Numbers of Arithmetic Theorems Not Definable in Arithmetic (and can be seen to follow immediately). {{qed}} {{Namedfor|Alfred Tarski|cat = Tarski}}
=== $({1}) \iff ({2})$ === The closure of $H$ is defined as the union of the set of all isolated points of $H$ and the set of all limit points of $H$. The rest then follows directly from Equivalence of Definitions of Isolated Point. {{qed|lemma}} === $({2}) \iff ({3})$ === Follows directly from Equivalence of Definitions of Adherent Point. {{qed|lemma}} === $({1}) \iff ({4})$ === The following equivalence holds: {{begin-eqn}} {{eqn | o =       | c = There exists an open neighborhood $U$ of $x$ such that $H \cap \paren {U \setminus \set x} = \O$ }} {{eqn | o = \leadstoandfrom       | c = There exists an open neighborhood $U$ of $x$ such that $U \subseteq H^{\complement} \cup \set x$       | cc = Modus Ponendo Tollens }} {{eqn | o = \leadstoandfrom       | c = $H^{\complement} \cup \set x$ is a neighborhood of $x$       | cc = {{Defof|Neighborhood of Point}} }} {{end-eqn}} The result follows from the Rule of Transposition. {{qed}} {{refactor|implement $({1}) \iff ({4})$ in house style}} === $({1}) \iff ({4})$, Proof Variant === The following equivalence holds: There exists an open neighborhood $U$ of $x$ such that $H \cap \paren {U \setminus \set x} = \O$ {{begin-eqn}} {{eqn | l = \O       | o = =       | r = H \cap \paren {U \setminus \set x} }} {{eqn | ll = \leadstoandfrom       | l  = \O       | o  = =       | r  = \paren {U \cap H} \setminus \set x       | cc = Intersection with Set Difference is Set Difference with Intersection }} {{eqn | ll = \leadstoandfrom       | l  = \O       | o  = =       | r  = \paren {H \cap U} \setminus \set x       | cc = Intersection is Commutative }} {{eqn | ll = \leadstoandfrom       | l  = \O       | o  = =       | r  = U \cap \paren {H \setminus \set x}       | cc = Intersection with Set Difference is Set Difference with Intersection }} {{eqn | ll = \leadstoandfrom       | l  = \O       | o  = =       | r  = U \cap \map \complement { \map \complement {H \setminus \set x} }       | cc = Complement of Complement }} {{eqn | ll = \leadstoandfrom       | l  = U       | o  = \subseteq       | r  = \map \complement {H \setminus \set x}       | cc = Intersection with Complement is Empty iff Subset }} {{eqn | ll = \leadstoandfrom       | l  = U       | o  = \subseteq       | r  = \map \complement {H \cap \map \complement {\set x} }       | cc = Set Difference as Intersection with Complement }} {{eqn | ll = \leadstoandfrom       | l  = U       | o  = \subseteq       | r  = \map \complement H \cup \map \complement {\map \complement {\set x} }       | cc = De Morgan's Laws (Set Theory)/Set Complement }} {{eqn | ll = \leadstoandfrom       | l  = U       | o  = \subseteq       | r  = \map \complement H \cup \set x       | cc = Complement of Complement }} {{end-eqn}} By {{Defof|Neighborhood of Point}}, $\map \complement H \cup \set x$ is a neighborhood of $x$ The result follows from the Rule of Transposition. {{qed|lemma}}
Let $H \subseteq S$ be everywhere dense in $T$.  As $\tau'$ is coarser than $\tau$, by definition $\tau' \subseteq \tau$. Let $x \in S$. As $H \subseteq S$ is everywhere dense in $T$, either $x \in H$ or $x$ is a limit point of $H$ in $T$. Suppose $x \notin H$. Then $x$ is a limit point of $H$ in $T$. Then by definition of limit point of $H$: :$\forall U \in \tau, x \in U: \exists y \in H: y \in U, y \ne x$  As $\tau' \subseteq \tau$ we have that $U \in \tau' \implies U \in \tau$. So: :$\forall U \in \tau', x \in U: \exists y \in H: y \in U, y \ne x$  That is, $x$ is a limit point of $H$ in $T\ '$. As $x$ is arbitrary, this follows for all $x \in S, x \notin H$. So, for all $x \in S$, either $x \in H$ or $x$ is a limit point of $H$ in $T'$. Thus, by definition, $H \subseteq S$ is everywhere dense in $T'$. {{qed}} Category:Denseness js5kepz4ph9o4ylhmlhwnvpvqnnonr1
We have that the Compact Complement Topology is Coarser than Euclidean Topology. The result follows from Denseness Preserved in Coarser Topology. {{qed}}
We have: : Compact Complement Topology is Second-Countable : Second-Countable Space is Separable Hence the result. {{qed}}
We have that: : Rationals are Dense in Compact Complement Topology : Rational Numbers are Countably Infinite Hence the result by definition of separable space. {{qed}}
:Either-Or Topology is Locally Path-Connected :Locally Path-Connected Space is Locally Connected {{qed}}
Let $\BB$ be a basis for $\tau$. Suppose that $0 \in B$, where $B \in \BB$. Then by definition of the either-or topology, $\openint {-1} 1 \subseteq B$. In particular, $\dfrac 1 2 \in B$. Let $f: \closedint 0 1 \to B$ be any injection such that: :$\map f 0 = 0$ :$\map f 1 = \dfrac 1 2$ Since $0 \notin \set {\dfrac 1 2}$, it follows that $\set {\dfrac 1 2}$ is open in $\tau$. As $f$ is an injection, it follows that: :$f^{-1} \sqbrk {\set {\dfrac 1 2} } = \set 1$ By Closed Real Interval is not Open Set, $\set 1 = \closedint 1 1$ is not open in $\closedint 0 1$. Thus, $f$ cannot be continuous. Hence, no arc from $0$ to $\dfrac 1 2$ can exist. It follows that $B$ is not arc-connected, and hence $T$ is not locally arc-connected. {{qed}}
Suppose $y \in S, y \ne p$. We have by definition of Fort space that $\set y$ is open in $T$. So there is no $z \in \set y: z \ne y, z \in U$.  Hence $y$ can not be a limit point of $U$. Suppose $p \in V \in \tau_p$ for some $V \subseteq S$. From {{Defof|Fort Space}}, $\relcomp S V$ is finite. Then: {{begin-eqn}} {{eqn | l = \relcomp S {V \setminus \set p}       | r = \paren {\relcomp S V} \cup \paren {V \cap \set p}       | c = Set Difference with Set Difference is Union of Set Difference with Intersection }} {{eqn | o = \subseteq        | r = \paren {\relcomp S V} \cup \set p       | c = Set Union Preserves Subsets: $V \cap \set p \subset \set p$: Intersection is Subset }} {{end-eqn}} Thus $\relcomp S {V \setminus \set p}$ is also finite. Since $U$ is countably infinite: :$U \not \subseteq \relcomp S {V \setminus \set p}$. From Empty Intersection iff Subset of Complement: :$U \cap \paren {V \setminus \set p} \ne \O$. Since $V$ is arbitrary, every open neighborhood of $p$ has non-empty intersection with $U$. Thus $p$ is the only limit point of $U$. {{qed}}
Let $H \subseteq T$ such that $H \ne \varnothing$ and $H \ne \set p$. Then $\exists x \in H: x \ne p$. From Clopen Points in Fort Space, every point of $T$ apart from $p$ is open in $T$. So $\set x$ is an open set of $T$. So $H \cap \set x = \set x$ and so $x$ is isolated in $H$. Thus $H$ contains at least one point which is isolated in $H$. On the other hand, suppose $H = \set p$. From Singleton Point is Isolated, $p$ is an isolated point in $\set p = H$. So again $H$ contains at least one point which is isolated in $H$. So for all $H \subseteq S$ we have that $H$ contains at least one point which is isolated in $H$. Hence the result, by definition of scattered space. {{qed}}
Suppose that $H \subseteq T$ has no isolated points of $H$. So, by definition, $H$ is dense in itself. We have that: :a Fort Space is $T_1$ :a Dense-in-itself Subset of $T_1$ Space is Infinite. So $H$ is infinite, and so contains more than one point. So $\exists q \in H: q \ne p$. But, from Clopen Points in Fort Space, $\set q$ is open in $T$. So $H \cap \set q = \set q$ and so by definition $q$ is isolated in $H$. From this contradiction it follows that $H$ is not dense in itself and contains at least one isolated point. Hence the result, by definition of scattered space. {{qed}}
Proof by Contradiction: {{AimForCont}} $H$ is finite. From Finite $T_1$ Space is Discrete, $H$ has the discrete topology. From Discrete Space is not Dense-In-Itself it then follows that $H$ can not be dense-in-itself. So for $H$ to be dense-in-itself, it must be infinite. {{qed}}
Suppose it is possible for two circles to touch at more points than one. :400px First, let the circle $ABDC$ touch the circle $EBFD$ internally at more than one point, that is, at $B$ and $D$. Let $G$ be the center of the circle $ABDC$, and $H$ be the center of the circle $EBFD$. (It is clear that in the diagram these centers are ''not'' actually at $G$ and $H$, and in fact $EBFD$ is obviously not a circle - it is the point of this proof to demonstrate that this would ''not'' be possible.) From Line Joining Centers of Two Circles Touching Internally the straight line $GH$ will pass through both $B$ and $D$. Since $G$ is the center of the circle $ABDC$, we have: {{begin-eqn}} {{eqn | l = BG       | r = GD       | c = }} {{eqn | ll= \implies       | l = BG       | o = >       | r = HD       | c = }} {{eqn | ll= \implies       | l = BH       | o = \gg       | r = HD       | c = }} {{end-eqn}} But since $H$ is the center of the circle $EBFD$, we have that $BH = HD$. But we have just shown that $BH \gg HD$, which is impossible. Therefore a circle does not touch another circle internally at more than one point. Next suppose the circle $ACK$ touches the circle $ABDC$ at more points than one, that is, at $A$ and $C$, and join $AC$. The two points $A$ and $C$ fall on the circumference of both circles $ABDC$ and $ACK$. So it follows from Chord Lies Inside its Circle that $AC$ lies with both circles. But from {{EuclidDefLink|III|3|Tangent Circles}} this line would fall inside $ABDC$ and outside $ACK$. This reveals a contradiction. Therefore a circle does not touch another circle externally at more than one point. {{qed}} {{Euclid Note|13|III}}
:300px Let $ABDC$ be a circle, and let $AB$ and $CD$ be equal chords on it. From Finding Center of Circle, let $E$ be the center of $ABDC$. Construct $EF$, $EG$ perpendicular to $AB$ and $CD$, and join $AE$ and $EC$. From Conditions for Diameter to be Perpendicular Bisector, $EF$ bisects $AD$. So $AF = FB$ and so $AB = 2 AF$. Similarly $CD = 2CG$. As $AB = CD$ it follows that $AE = CG$. Since $AE = EC$ the square on $AE$ equals the square on $EC$. We have that $\angle AFE$ is a right angle. So from Pythagoras's Theorem the square on $AE$ equals the sum of the squares on $AF$ and $FE$. Similarly, the square on $EC$ equals the sum of the squares on $EG$ and $GC$. Thus the square on $EF$ equals the square on $EG$. So $EF = EG$. So from Definition III: 4, chords are at equal distance from the center when the perpendiculars drawn to them from the center are equal. So $AB$ and $CD$ are at equal distance from the center. {{qed|lemma}} Now suppose that $AB$ and $CD$ are at equal distance from the center. That is, let $EF = EG$. Using the same construction as above, it is proved similarly that $AB = 2AF$ and $CD = 2CG$. Since $AE = CE$, the square on $AE$ equals the square on $CE$. From Pythagoras's Theorem: : the square on $AE$ equals the sum of the squares on $AF$ and $FE$ : the square on $EC$ equals the sum of the squares on $EG$ and $GC$. So the sum of the squares on $AF$ and $FE$ equals the sum of the squares on $EG$ and $GC$. But the square on $FE$ equals the square on $EG$. So the square on $AF$ equals the square on $CG$, and so $AF = CG$. As $AB = 2AF$ and $CD = 2CG$ it follows that $AB = CD$. Hence the result. {{qed}} {{Euclid Note|14|III}}
:300px Let $ABCD$ be a clrcle, let $AD$ be its diameter and $E$ the center. Let $BC$ and $FG$ be chords of $ABCD$, where $BC$ is nearer to the center than $FG$. Let $EH$ and $EK$ be drawn perpendicular to $BC$ and $FG$ respectively. Because $BC$ is nearer to the center than $FG$, it follows from Book III: Definition 5 that $EK$ is longer than $EH$. Position $L$ on $EK$ so that $EL = EH$, and draw $MN$ through $L$ perpendicular to $FG$. Join $EM$, $EN$, $EF$ and $EG$. Since $EH = EL$ it follows from Equal Chords in Circle that $BC = MN$. Also, since $AE = EM$ and $ED = EN$, $AD = ME + EN$. But $ME + EN > MN$ from Sum of Two Sides of Triangle Greater than Third Side, and $MN = BC$. So $AD > BC$. Thus we have that the diameter is greater than any other chord of a circle. {{qed|lemma}} Since $ME = FE$ and $EN = EG$ and $\angle MEN > \angle FEG$, from the Hinge Theorem it follows that $MN > FG$. But $MN = BC$. So a chord nearer the center is greater than one more remote. {{qed}} {{Euclid Note|15|III}}
:400px Let $ABC$ be a circle about $D$ as center and $AB$ as diameter. Suppose that a line from $A$ at right angles to the diameter falls within the circle, e.g. at $AC$. Let $DC$ be joined. Since $DA = DC$ we have that $\angle DAC = \angle ACD$. But by hypothesis the angle $DAC$ is a right angle. Therefore $\angle ACD$ is also a right angle. So in $\triangle ACD$, the two angles $\angle DAC$ and $\angle ACD$ equal two right angles. From Two Angles of Triangle Less than Two Right Angles this is impossible. Therefore the straight line drawn at right angles to the diameter of a circle from its extremity will fall outside the circle. {{qed|lemma}} Let this line be as $AE$. Suppose there were another straight line $AF$ interposed between the circumference of the circle and $AE$. Let $DG$ be drawn perpendicular to $AF$ from $D$. Since $\angle AGD$ is a right angle, the angle $\angle DAG$ is less than a right angle by Two Angles of Triangle Less than Two Right Angles. So $AD > DG$ from Greater Angle of Triangle Subtended by Greater Side. But $AD = DH$ and so $DH > DG$, which is impossible. Therefore another straight line cannot be interposed into the space between the straight line and the circumference of the circle. {{qed|lemma}} Suppose there were a rectilineal angle greater than: : that contained by the straight line $AB$ and the arc $CHA$ of the circle and: : any rectilineal angle less than the angle contained by the arc $CHA$ and the straight line $AE$. Then into the space between the circumference and the straight line $AE$ another straight line can be interposed such as will make an angle contained by straight lines greater than the angle contained by the straight line $AB$ and the arc $CHA$. Also, another angle contained by the straight lines $AB$ and $AE$ which is less than the angle contained by the arc $CHA$ and the straight line $AE$. But such a straight line cannot be interposed. Therefore there will not be any acute angle contained by straight lines which is greater than the angle contained by the straight line $AB$ and the arc $CHA$. Nor will there be any any acute angle contained by straight lines which is less than the angle contained by the arc $CHA$ and the straight line $AE$. {{qed}} {{Euclid Note|16|III}}
:400px Let $A$ be the given point and let $BCD$ be the given circle. It is required that a straight line be drawn from $A$ to $BCD$. Let the center $E$ of $BCD$ be found. Join $AE$ and draw the circle $AFG$ with center $E$ and radius $AE$. From $D$ let $DF$ be drawn perpendicular to $EA$. Join $EF$ and let $B$ be the point at which $EF$ joins the circle $BCD$. Join $AB$. Then $AB$ is the required tangent to $BCD$. === Proof of Construction === Since $E$ is the center of the circles $BCD$ and $AFG$, $EA = EF$ and $ED = EB$. Therefore the two sides $AE, EB$ equal the two sides $FE, ED$, and they contain a common angle at $E$. So by Triangle Side-Angle-Side Equality, $\triangle DEF = \triangle BEA$. Therefore $\angle EDF = \angle EBA$. But $\angle EDF$ is a right angle, so $\angle EBA$ is also a right angle. But $EB$ is a radius of circle $BCD$. It follows from the porism to Line at Right Angles to Diameter of Circle that $AB$ is tangent to $BCD$. {{qed}}
:400px Let $BCD$ with center $A$ be the circle, and let $E$ be the exterior point from which a tangent is to be drawn. Bisect $AE$ at $F$. Then draw a circle $AEG$ whose center is $F$ and whose radius is $AF$. The point $G$ is where $AEG$ intersects $BCD$. The line $EG$ is the required tangent. === Proof of Construction === By the method of construction, $AE$ is the diameter of $AEG$. By Thales' Theorem $\angle AGE$ is a right angle. But $AG$ is a radius of $BCD$. The result follows from Radius at Right Angle to Tangent. {{qed}}
:400px Let $DE$ be tangent to the circle $ABC$ at $C$. Let $F$ be the center of $ABC$. Let $FC$ be the radius in question. Suppose $FC$ were not perpendicular to $DE$. Instead, suppose $FG$ were drawn perpendicular to $DE$. Since $\angle FGC$ is a right angle, then from Two Angles of Triangle Less than Two Right Angles it follows that $\angle FCG$ is acute. From Greater Angle of Triangle Subtended by Greater Side it follows that $FC > FG$. But $FC = FB$ and so $FB > FG$, which is impossible. So $FG$ can not be perpendicular to $DE$. Similarly it can be proved that no other straight line except $FC$ can be perpendicular to $DE$. Therefore $FC$ is perpendicular to $DE$. {{qed}} {{Euclid Note|18|III}}
:400px Let $DE$ touch the circle $ABC$, and let $AC$ be drawn at right angles to $DE$. Suppose the center is not on $AC$. Let $F$ be the center of $ABC$, and let $FC$ be joined from there to the point of contact. By Radius at Right Angle to Tangent, $FC$ is perpendicular to $DE$. Therefore $\angle FCE$ is a right angle. But by hypothesis $\angle ACE$ is also a right angle. Thus $\angle FCE = \angle ACE$, which is impossible unless $F$ lies on $AC$. Therefore the center of $ABC$ lies on $AC$. {{Qed}} {{Euclid Note|19|III}}
:400px Let $ABCD$ be a circle, and let $\angle BAD, \angle BED$ be angles in the same segment $BAED$. Let $F$ be the center of $ABCD$, and join $BF$ and $FD$. From the Inscribed Angle Theorem: :$\angle BFD = 2 \angle BAD$ :$\angle BFD = 2 \angle BED$ So $\angle BAD = \angle BED$. Hence the result. {{qed}} {{Euclid Note|21|III}}
:350px Let $ABCD$ be a cyclic quadrilateral in a circle. Join $AC$ and $BD$. From Sum of Angles of Triangle equals Two Right Angles, we have that $\angle CAB + \angle ABC + \angle CBA$ equals two right angles. But from Angles in Same Segment of Circle are Equal we have that $\angle CAB = \angle BDC$ and also $\angle ACB = \angle ADB$. If we add $\angle ABC$ to each, we get that $\angle ABC + \angle BAC + \angle ACB = \angle ABC + \angle ADC$. Therefore $\angle ABC + \angle ADC$ equals two right angles. Similarly we can show that $\angle BAD + \angle DCB$ also equals two right angles. {{Qed}} {{Euclid Note|22|III}}
:300px Suppose it were possible to construct two similar and unequal segments $ACB$ and $ADB$ on the same base $AB$. Let $ACD$ be drawn through, and join $CB$ and $CB$. We have by hypothesis that segment $ACD$ is similar to $ADB$. We also have by {{EuclidDefLink|III|11|Similar Segments}} that similar segments admit equal angles. So $\angle ACB = \angle ADB$, which from External Angle of Triangle Greater than Internal Opposite is impossible. {{qed}} {{Euclid Note|23|III}}
:600px Let $AEB$ and $CFD$ be similar segments of circles on equal bases $AB$ and $CD$. Let the segment $AEB$ be applied to $CFD$ so that $A$ be placed on $C$ and $AB$ on $CD$. Then $B$ will coincide with $D$ as $AB = CD$. Suppose that segment $AEB$ does not coincide with segment $CFD$. It will fall in one of three ways: :$(1) \quad$ Inside it :$(2) \quad$ Outside it :$(3) \quad$ Awry, as $CGD$. If $CFD$ falls inside or outside $AEB$, then by definition $AEB$ and $CFD$ are not similar. But from Two Circles have at most Two Points of Intersection option $(3)$ is impossible. Hence the result. {{Qed}} {{Euclid Note|24|III}}