hcju/proofwikiqa huggingface dataset:
Let $ABC$ be the given segment of a circle whose base is $AC$. Bisect $AC$ at $D$, draw $DB$ perpendicular to $AC$, and join $AB$. First suppose that $ABC$ is such that $\angle ABD > \angle BAD$: :250px On $BA$, construct $\angle BAE$ equal to $\angle ABD$. Join $BD$ through to $E$ and join $EC$. Then $E$ is the center of the required circle. Second, suppose that $ABC$ is such that $\angle ABD = \angle BAD$: :250px Then $D$ is the center of the required circle. Finally, suppose that $ABC$ is such that $\angle ABD < \angle BAD$: :250px On $BA$, construct $\angle BAE$ equal to $\angle ABD$. The point $E$, which falls on $BD$, is the center of the required circle. === Proof of Construction === First suppose that $ABC$ is such that $\angle ABD > \angle BAD$: :250px Since $\angle ABE = \angle BAE$, from Triangle with Two Equal Angles is Isosceles we have that $EB = EA$. Since $AD = DC$ and $DE$ is common, and $\angle ADE$ is a right angle, by Triangle Side-Angle-Side Equality we have that $\triangle ADE = \triangle CDE$. Hence $AE = CE$ both of which are equal to $BE$ from above. So from Condition for Point to be Center of Circle $E$ is the center of the required circle. Second, suppose that $ABC$ is such that $\angle ABD = \angle BAD$: :250px From Triangle with Two Equal Angles is Isosceles we have that $AD = DB$ and so also equal to $DC$. So from Condition for Point to be Center of Circle $D$ is the center of the required circle. Incidentally, note that in this case segment $ABC$ is actually a semicircle. Finally, suppose that $ABC$ is such that $\angle ABD < \angle BAD$: :250px The same proof applies: Since $\angle ABE = \angle BAE$, from Triangle with Two Equal Angles is Isosceles we have that $EB = EA$. Since $AD = DC$ and $DE$ is common, and $\angle ADE$ is a right angle, by Triangle Side-Angle-Side Equality we have that $\triangle ADE = \triangle CDE$. Hence $AE = CE$ both of which are equal to $BE$ from above. So from Condition for Point to be Center of Circle $E$ is the center of the required circle. {{qed}}
:250px Choose any point $C$ on the circumference. Bisect $AC$ at $D$ and $BC$ at $E$ and construct a perpendicular $DF$ and $EF$ from each through the point of bisection. The point of intersection $F$ is the center of the required circle. $AFC$ and $CFB$ are isosceles triangles and so $AF, CF$ and $BF$ are all equal. The result follows from Condition for Point to be Center of Circle. {{qed}}
Let $ABC$ and $DEF$ be equal circles. Let $\angle BGC = \angle EHF$ and $\angle BAC = \angle EDF$. :500px Let $BC$ and $EF$ be joined. Since the circles $ABC$ and $DEF$ are equal, their radii are equal. So $BG = EH$ and $CG = FH$. We also have by hypothesis that $\angle BGC = \angle EHF$. So from Triangle Side-Angle-Side Equality it follows that $BC = EF$. Since $\angle BAC = \angle EDF$ we have from {{EuclidDefLink|III|11|Similar Segments}} that segment $BAC$ is similar to segment $EDF$. Moreover, these segments have equal bases. So from Similar Segments on Equal Bases are Equal, segment $BAC$ is equal to segment $EDF$.  But as $ABC$ and $DEF$ are equal circles, it follows that arc $BKC$ equals arc $ELF$. {{qed}} {{Euclid Note|26|III|{{EuclidNoteConverse|prop=27|title=Angles on Equal Arcs are Equal}}}}
Let $ABC$ and $DEF$ be equal circles. Let $\angle BGC$ and $\angle EHF$ stand on the equal arcs $BC$ and $EF$. :500px Suppose $\angle BGC \ne \angle EHF$. Then one of them is bigger. Suppose $\angle BGC > \angle EHF$. On the straight line $BG$, construct $\angle BGK$ equal to $\angle EHF$. From Equal Angles in Equal Circles, these angles stand on equal arcs. So arcs $BK = EF$. But $EF = BC$ and so $BK = BC$, which is impossible. So $\angle BGC$ and $\angle EHF$ are not unequal, therefore $\angle BGC = \angle EHF$. From the Inscribed Angle Theorem, $2 \angle BAC = \angle BGC$ and $2 \angle EDF = \angle EHF$. Therefore $2 \angle BAC = \angle EHF$. {{qed}} {{Euclid Note|27|III|{{EuclidNoteConverse|prop=26|title=Equal Angles in Equal Circles}}}}
Let $ABC$ and $DEF$ be equal circles. Let $AB, DE$ be equal straight lines cutting off arcs $ACB$ and $DFE$ as the greater, and $AGB, DHE$ as lesser. :500px Let $K$ and $L$ be the centers of the circles $ABC$ and $DEF$ respectively. Let $AK, KB, DL, LE$ be joined. Since the circles are equal, so are their radii. So $AK = DL, KB = LE$. As $AB = DE$ by hypothesis, from Triangle Side-Side-Side Equality it follows that $\angle AKB = \angle DLE$. But from Angles on Equal Arcs are Equal the arc $AGB$ equals the arc $DHE$. As the whole circles $ABC$ and $DEF$ are equal, the arc $ACB$ which remains also equals arc $DFE$. {{qed}} {{Euclid Note|28|III|{{EuclidNoteConverse|prop=29|title=Equal Arcs of Circles Subtended by Equal Straight Lines}}}}
Let $ABC$ and $DEF$ be equal circles. Let equal arcs $BGC$ and $EHF$ be cut off by the straight lines $BC$ and $EF$. :500px Let $K$ and $L$ be the centers of the circles $ABC$ and $DEF$ respectively. Let $BK, KC, EL, LF$ be joined. We have that the arcs $BGC$ and $EHF$ are equal. So from Angles on Equal Arcs are Equal $\angle BKC = \angle ELF$. Since circles $ABC$ and $DEF$ are equal, so are their radii. So $BK = EL$ and $KC = LF$, and they contain equal angles. So from Triangle Side-Angle-Side Equality base $BC$ is equal to base $EF$. {{qed}} {{Euclid Note|29|III|{{EuclidNoteConverse|prop=28|title=Straight Lines Cut Off Equal Arcs in Equal Circles}}}}
Join $AD$ and $BD$. We have that $AC = CB$ and $CD$ is common. We also have that $\angle ACD = \angle BCD$ as they are both right angles. So from Triangle Side-Angle-Side Equality $\triangle ACD = \triangle BCD$ and so $AD = BD$. But from Straight Lines Cut Off Equal Arcs in Equal Circles, the arc $AD$ equals the arc $BD$. That is, the arc $AB$ has been bisected at $D$. {{qed}} {{Euclid Note|30|III}}
:300px Let $ABCD$ be a circle whose diameter is $BC$ and whose center is $E$. Join $AB$, $AC$, $AD$, $DC$ and $AE$. Let $BA$ be produced to $F$. Since $BE = EA$, from Isosceles Triangle has Two Equal Angles it follows that $\angle ABE = \angle BAE$. Since $CE = EA$, from Isosceles Triangle has Two Equal Angles it follows that $\angle ACE = \angle CAE$. So from $\angle BAC = \angle ABE + \angle ACE = \angle ABC + \angle ACB$. But from Sum of Angles of Triangle Equals Two Right Angles $\angle FAC = \angle ABC + \angle ACB$. So $\angle BAC = \angle FAC$, and so from Book I Definition 10 each one is a right angle. So the angle in the semicircle $BAC$ is a right angle. {{qed|lemma}} From Two Angles of Triangle Less than Two Right Angles, in $\triangle ABC$, $\angle ABC + \angle BAC$ is less than two right angles. As $\angle BAC$ is a right angle, it follows that $\angle ABC$ is less than a right angle. It is also the angle in a segment $ABC$ greater than a semicircle. Therefore the angle in a segment greater than a semicircle is acute. {{qed|lemma}} We have that $ABCD$ is a cyclic quadrilateral. From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $\angle ABC + \angle ADC$ equals two right angles. As $\angle ABC$ is less than a right angle, it follows that $\angle ADC$ is greater than a right angle. But $\angle ADC$ is the angle in a segment $ADC$ less than a semicircle. Therefore the angle in a segment less than a semicircle is obtuse. {{qed|lemma}} Since the angle contained by the straight lines $BA$ and $AC$ is a right angle, the angle contained by the arc $ABC$ and the straight line $AC$ is greater than a right angle. So the angle of a segment greater than a semicircle is obtuse. {{qed|lemma}} Since the angle contained by the straight lines $AC$ and $AF$ is a right angle, the angle contained by the arc $ADC$ and the straight line $AC$ is less than a right angle. So the angle of a segment less than a semicircle is acute. {{qed}} {{Euclid Note|31|III}}
:250px Draw $BA$ perpendicular to $EF$ through $B$. Let $C$ be selected on the circle on the arc $BD$. Join $AD, DC, CB$. From Right Angle to Tangent of Circle goes through Center, the center of the circle lies on $AB$. By definition, then, $AB$ is a diameter of the circle. From Relative Sizes of Angles in Segments, it follows that $\angle ADB$ is a right angle. Therefore from Sum of Angles of Triangle Equals Two Right Angles $\angle BAD + \angle ABD$ equals a right angle. But $\angle ABF$ is also a right angle. So $\angle ABF = \angle BAD + \angle ABD$ Subtracting $\angle ABD$ from each, it follows that $\angle DBF = \angle BAD$. {{qed|lemma}} Next we have that $ABCD$ is a cyclic quadrilateral. From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles $\angle BAD + \angle BCD$ equals two right angles. But from Two Angles on Straight Line make Two Right Angles, so does $\angle DBE + \angle DBF$. But as $\angle BAD = \angle DBF$ it follows that $\angle BCD = \angle DBE$. {{qed}} {{Euclid Note|32|III}}
=== Proof for Acute Angle === We have that $AF = FB$, $\angle AFG = \angle BFG$ (both are right angles) and $AG$ is common. So from Triangle Side-Angle-Side Equality it follows that $AG = BG$. So the circle $ABE$ whose center is at $G$ and whose radius is $GB$ also passes through $A$. As $AE$ passes through the  center $G$, $AE$ is a diameter of the circle $ABE$. As $AD$ is perpendicular to $AE$, it follows from Line at Right Angles to Diameter of Circle that $AD$ is tangent to circle $ABE$. From Angles made by Chord with Tangent, $\angle DAB = \angle AEB$. But $\angle DAB = \angle C$ by construction. Hence the result. {{qed|lemma}} === Proof for Right Angle === Select a point $E$ on the circle and join $AE$ and $BE$. As $AD$ is perpendicular to $AB$, it follows from Line at Right Angles to Diameter of Circle that $AD$ is tangent to circle $ABE$. From Relative Sizes of Angles in Segments, $\angle BAD = \angle AEB$. But as $\angle BAD = \angle C$ by construction, $\angle AEB = \angle C$. Hence the result. {{qed|lemma}} === Proof for Obtuse Angle === We have that $AF = FB$, $\angle AFG = \angle BFG$ (both are right angles) and $AG$ is common. So from Triangle Side-Angle-Side Equality it follows that $AG = BG$. So the circle $ABE$ whose center is at $G$ and whose radius is $GB$ also passes through $A$. As $AE$ passes through the  center $G$, $AE$ is a diameter of the circle $ABE$. As $AD$ is perpendicular to $AE$, it follows from Line at Right Angles to Diameter of Circle that $AD$ is tangent to circle $ABE$. Now place point $H$ anywhere on the arc $AB$ and join $AH$ and $HB$. From Angles made by Chord with Tangent, $\angle DAB = \angle AHB$. But $\angle DAB = \angle C$ by construction. Hence the result. {{qed}} {{Euclid Note|33|III|Note that Euclid, in his original constructions for the acute and obtuse angle, defined the line perpendicular to $AD$ as being $AE$. But at that stage of the construction, it has not been established exactly where $E$ is going to be. This is defined after the position of $G$ on $AB$ has been established, and the circle centered at $G$ has been drawn.<br /><br />It is also worth pointing out that while the constructions for the acute and obtuse angles are the same, the proofs need to be done separately for each.}}
Let $A$ be selected anywhere on the circle opposite the segment in question. From Angles made by Chord with Tangent, $\angle BAC = \angle FBC$. But $\angle FBC = \angle D$. Hence the result. {{qed}} {{Euclid Note|34|III}}
Let $DA$ pass through the center $F$ of circle $ABC$. Join $FB$. From Radius at Right Angle to Tangent, $\angle FBD$ is a right angle. :250px We have that $F$ bisects $AC$ and that $CD$ is added to it. So we can apply Square of Sum less Square and see that: :$AD \cdot DC + FC^2 = FD^2$ But $FC = FB$ and so: :$AD \cdot DC + FB^2 = FD^2$ But from Pythagoras's Theorem we have that $FD^2 = FB^2 + DB^2$ and so: :$AD \cdot DC + FB^2 = FB^2 + DB^2$ from which it follows that: :$AD \cdot DC = DB^2$ which is what we wanted to show. {{qed|lemma}} Now let $DA$ be such that it does not pass through the center $E$ of circle $ABC$. Draw $EF$ perpendicular to $DA$ and draw $EB, EC, ED$. :250px From Radius at Right Angle to Tangent, $\angle EBD$ is a right angle. From Conditions for Diameter to be Perpendicular Bisector, $EF$ bisects $AC$. So $AF = FC$. So we can apply Square of Sum less Square and see that: :$AD \cdot DC + FC^2 = FD^2$ Let $FE^2$ be added to each: :$AD \cdot DC + FC^2 + FE^2 = FD^2 + FE^2$ Now $\angle DFE$ is a right angle and so by Pythagoras's Theorem we have: : $FD^2 + FE^2 = ED^2$ : $FC^2 + FE^2 = EC^2$ This gives us: :$AD \cdot DC + EC^2 = ED^2$ But $EC = EB$ as both are radii of the circle $ABC$. Next note that $\angle EBD$ is a right angle and so by Pythagoras's Theorem we have: :$ED^2 = EB^2 + DB^2$ which gives us: :$AD \cdot DC + EB^2 = EB^2 + DB^2$ from which it follows that: :$AD \cdot DC = DB^2$ which is what we wanted to show. {{qed}} {{Euclid Note|36|III|{{EuclidNoteConverse|prop=37|title=Converse of Tangent Secant Theorem}}}}
Firstly, by definition of multiplication: {{begin-eqn}} {{eqn | l = n \times 1       | r = \paren {n \times 0} + n }} {{eqn | r = n }} {{end-eqn}} Next, recall that multiplication is recursively defined as: :$\forall m, n \in \N: \begin{cases} m \times 0 & = 0 \\ m \times \paren {n + 1} & = m \times n + m \end{cases}$ From the Principle of Recursive Definition, there is only one mapping $f$ satisfying this definition for $m = 1$; that is, such that: :$\forall n \in \N: \begin {cases} \map f 0 = 0 \\ \map f {n + 1} = \map f n + 1 \end{cases}$ Consider now $f'$ defined as $\map {f'} n = n$. Then evidently $\map {f'} 0 = 0$. Also: {{begin-eqn}} {{eqn | l = \map {f'} {n + 1}       | r = n + 1 }} {{eqn | r = \map {f'} n + 1 }} {{end-eqn}} showing that $f'$ also satisfies the definition for $1 \times n$. Hence $n \times 1 = 1 \times n = n$ for all $n \in \N$. That is, $1$ is the identity element for multiplication. {{qed}}
:300px Draw $CF$ tangent to the circle. From the Tangent Secant Theorem we have that: :$CF^2 = CA \cdot CD$ :$CF^2 = CB \cdot CE$ from which the result is obvious and immediate. {{qed}}
:300px Let $DE$ be drawn tangent to the circle $ABC$. Let $F$ be the center of $ABC$ and join $FB, FD, FE$. From Radius at Right Angle to Tangent, $\angle FED$ is a right angle. We have that $DE$ is tangent to the circle $ABC$ and $DA$ cuts it. So from the Tangent Secant Theorem $AD \cdot DC = DE^2$. But we also have by hypothesis that $AD \cdot DC = DB^2$. So $DE^2 = DB^2$ and so $DE = DB$. Also $FE = FB$ and so $DE, EF = DB, DF$ and $DF$ is common. So from Triangle Side-Side-Side Equality $\angle DEF = \angle DBF$. But $\angle DEF$ is a right angle and so also is $\angle DBF$. So from Line at Right Angles to Diameter of Circle $DB$ is tangent to the circle $ABC$. {{qed}} {{Euclid Note|37|III|{{EuclidNoteConverse|prop=36|title=Tangent Secant Theorem}}}}
As $C$ is the center of $EFG$, $CG = CE$ and so $CG = D$. But $CG$ is a chord of $ABC$. {{Qed}} {{Euclid Note|1|IV}}
We have that $AH$ touches the circle $ABC$, and $AC$ is a chord of circle $ABC$. Then from Angles made by Chord with Tangent $\angle HAC = \angle ABC$. But $\angle HAC = \angle DEF$. Similarly, we have that $AG$ touches the circle $ABC$, and $AB$ is a chord of circle $ABC$. Then from Angles made by Chord with Tangent $\angle GAB = \angle ACB$. But $\angle GAB = \angle DFE$. So from Sum of Angles of Triangle Equals Two Right Angles it follows that the remaining angles are equal also: $\angle BAC = \angle EDF$. Hence the result. {{qed}} {{Euclid Note|2|IV}}
From Radius at Right Angle to Tangent, $LM$ is perpendicular to $AK$, $MN$ is perpendicular to $BK$, and $LN$ is perpendicular to $CK$. We have that $\angle KAM, \angle KBM$ are right angles. This leaves us with $\angle AMB + \angle AKB$ equal to two right angles. We have that $\angle AKB = \angle DEG$ and so $\angle AMB + \angle DEG$ equals two right angles. But from Two Angles on Straight Line make Two Right Angles $\angle DEF + \angle DEG$ equals two right angles. It follows that $\angle AMB = \angle DEF$. In a similar way we can show that $\angle ALC = \angle EDF$ and $\angle BNC = \angle DFE$. Hence the result. {{qed}} {{Euclid Note|3|IV}}
We have that $\angle ABD = \angle CBD$ and $\angle BED = \angle BFD$, a right angle, and $BD$ is common. So from Triangle Angle-Side-Angle Equality we have that $\triangle EBD = \triangle FBD$. So $DE = DF$. For the same reason $DG = DF$. So $DE = DF = DG$. So the circle drawn with radius $DE$ will pass through $E, F$ and $G$. From Line at Right Angles to Diameter of Circle it follows that $AB, AC, BC$ are tangent to the circle $EFG$. Hence the result. {{qed}} {{Euclid Note|4|IV}}
The point $F$ will be either inside, outside or on the edge $BC$ of $\triangle ABC$. :250px First suppose $F$ is inside $\triangle ABC$. Join $FB$ and $FC$. We have that $AD = DB$ and $DF$ is common and at right angles to $AB$. So from Triangle Side-Angle-Side Equality, $\triangle ADF = \triangle BDF$, and so $AF = BF$. Similarly we can prove that $BF = CF$. So $AF = BF = CF$. Thus the circle with center $F$ and radius $AF$ also passes through $B$ and $C$. That is, it circumscribes $\triangle ABC$. {{qed|lemma}} :250px Secondly, suppose $F$ lies on $BC$. We have that $AD = DB$ and $DF$ is common and at right angles to $AB$. So from Triangle Side-Angle-Side Equality, $\triangle ADF = \triangle BDF$, and so $AF = BF$. Similarly we can prove that $BF = CF$. So $AF = BF = CF$. Thus the circle with center $F$ and radius $AF$ also passes through $B$ and $C$. That is, it circumscribes $\triangle ABC$. {{qed|lemma}} :250px Thirdly, suppose $F$ lies outside $\triangle ABC$. Join $FB$ and $FC$. We have that $AD = DB$ and $DF$ is common and at right angles to $AB$. So from Triangle Side-Angle-Side Equality, $\triangle ADF = \triangle BDF$, and so $AF = BF$. Similarly we can prove that $BF = CF$. So $AF = BF = CF$. Thus the circle with center $F$ and radius $AF$ also passes through $B$ and $C$. That is, it circumscribes $\triangle ABC$. {{qed}}
We have that $BE = ED$, $EA$ is common and $\angle BEA = \angle DEA$ are right angles. So from Triangle Side-Angle-Side Equality, $\triangle ABE = \triangle ADE$ and so $AB = AD$. For the same reason $BC = CD = AD$ and so all four sides $AB, BC, CD, DA$ are equal. So $\Box ABCD$ is equilateral. Next we have that $BD$ is a diameter of circle $ABCD$. So from Relative Sizes of Angles in Segments, $\angle BAD$ is a right angle. For the same reason, $\angle ABC$, $\angle BCD$ and $\angle ADC$ are also all right angles. So by definition, $\Box ABCD$ is a square. {{Qed}} {{Euclid Note|6|IV}}
From Line at Right Angles to Diameter of Circle, $FG, GH, HK, KF$ will be tangent to $ABCD$. Since $\angle AEB$ is a right angle, $\angle EBG$ is also a right angle. So $GH$ is parallel to $AC$. For the same reason $FK$ is parallel to $AC$. From Parallelism is Transitive Relation, $GH$ is parallel to $FK$. Similarly, each of $GF, HK$ is parallel to $BD$. So $GK, GC, AK, FB, BK$ are all parallelograms. Therefore from Opposite Sides and Angles of Parallelogram are Equal $GF = HK$ and $GH = FK$. We have that $AC = BD, AC = GH, AC = FK$. Also $BD = GF = HK$. So $\Box FGHK$ is equilateral. Since $GBEA$ is a parallelogram, $\angle AEB$ is a right angle. Therefore $\angle AGB$ is also a right angle. Similarly the angles at $H, K, F$ are also right angles. So $\Box FGHK$ is a square. {{qed}} {{Euclid Note|7|IV}}
Each of the figures $AK, KB, AH, FD, AG, GC, BG, GD$ is a parallelogram from the method of construction. The opposite sides of all of them is equal. We have that $AD = AB$ and $2 AE = AD$, and $2 AF = AB$. Therefore $AE = AF$. Because the opposite sides are equal, $FG = GE$. Similarly we can show that $GE = GF = GH = GK$. So the circle whose center is at $G$ and whose radius is $EG$ passes through all of $E, F, H, K$. The lines $AB, BC, CD, DA$ are all perpendicular to the diameters $FK$ and $EH$. So from Line at Right Angles to Diameter of Circle, these lines are tangent to the circle. Hence the result. {{qed}} {{Euclid Note|8|IV}}
We have that $DA = AB$, $DC = BC$ and $AC$ is common. So from Triangle Side-Side-Side Equality $\triangle DAB = \triangle DCB$ and so $\angle DAC = \angle BAC$. So $\angle DAB$ is bisected by $AC$. Similarly each of $\angle ABC, \angle BCD, \angle DCA$ are bisected by $AC$ or $DB$. We have that $\angle DAB = \angle ABC$, $2 \angle EAB = \angle DAB$, $2 \angle EBA = \angle ABC$. Then $\angle EAB = \angle EBA$. From Triangle with Two Equal Angles is Isosceles, $EA = EB$. Similarly, $EA = EB = EC = ED$. So the circle whose center is $E$ and whose radius is $AE$ passes through each of $B, C, D$. Hence the result. {{qed}} {{Euclid Note|9|IV}}
Join $CD$. Circumscribe circle $ACD$ about $\triangle ACD$. As $AC = BD$ we have that $AB \cdot BC = BD^2$. We have that $B$ is outside the circle $ACD$. From the converse of the Tangent Secant Theorem it follows that $BD$ is tangent to circle $ACD$. Then from Angles made by Chord with Tangent $\angle BDC = \angle DAC$. Add $\angle CDA$ to both: :$\angle CDA + \angle BDC = \angle BDA = \angle CDA + \angle DAC$. But from Sum of Angles of Triangle Equals Two Right Angles we have that: : $(1) \quad \angle BCD = \angle CDA + \angle DAC$ So $\angle BDA = \angle BCD$. But since $AD = AB$, from Isosceles Triangle has Two Equal Angles $\angle BDA = \angle CBD$. So $\angle BDA = \angle BCD = \angle CBD$. Since $\angle DBC = \angle BCD$, from Triangle with Two Equal Angles is Isosceles we have $BD = DC$. But by hypothesis $BD = CA$ and so $CA = CD$. So from Isosceles Triangle has Two Equal Angles $\angle CDA = \angle DAC$. So $\angle CDA + \angle DAC = 2 \angle DAC$. But from $(1)$ we have that $\angle BCD = \angle CDA + \angle DAC$. So $\angle BCD = 2 \angle CAD = 2 \angle BAD$. But $\angle BCD = \angle BDA = \angle DBA$. So $\angle ABD = \angle BAD = 2 \angle BDA$. {{qed}} {{Euclid Note|10|IV|Having established in the proof that $CD$ equals $BD$, the construction can be simplified by constructing the circle whose center is at $C$ and whose radius is $AC$, then identifying $D$ as the point at which circle $ACD$ meets circle $ABD$, instead of invoking the somewhat more cumbersome construction that fits $BD$ into the circle $ABD$.}}
We have that $\angle CDA = \angle DCA = 2 \angle CAD$. As $\angle CDA$ and $ \angle DCA$ have been bisected, $\angle DAC = \angle ACE = \angle ECD = \angle CDB = \angle BDA$. From Equal Angles in Equal Circles it follows that the arcs $AB, BC, CD, DE, EA$ are all equal. Hence from Equal Arcs of Circles Subtended by Equal Straight Lines, the straight lines $AB, BC, CD, DE, EA$ are all equal. So the pentagon $ABCDE$ is equilateral. Now since the arc $AB$ equals arc $DE$, we can add $BCD$ to each. So the arc $ABCD$ equals arc $BCDE$. So from Angles on Equal Arcs are Equal $\angle BAE = \angle AED$. For the same reason, $\angle BAE = \angle AED = \angle ABC = \angle BCD = \angle CDE$. So the pentagon $ABCDE$ is equiangular. {{qed}} {{Euclid Note|11|IV}}