hcju/proofwikiqa huggingface dataset:
Construct the circumcircle of $\triangle ABC$, let $O$ be the circumcenter and $R$ be the circumradius. Construct $\triangle AOB$ and let $E$ be the foot of the altitude of $\triangle AOB$ from $O$. :350px By the Inscribed Angle Theorem: :$\angle ACB = \dfrac {\angle AOB} 2$ From the definition of the circumcenter: :$AO = BO$ From the definition of altitude and the fact that all right angles are congruent: :$\angle AEO = \angle BEO$ Therefore from Pythagoras's Theorem: :$AE = BE$ and then from Triangle Side-Side-Side Equality: :$\angle AOE = \angle BOE$ Thus: :$\angle AOE = \dfrac {\angle AOB} 2$ and so: :$\angle ACB = \angle AOE$ Then by the definition of sine: :$\sin C = \map \sin {\angle AOE} = \dfrac {c / 2} R$ and so: :$\dfrac c {\sin C} = 2 R$ The same argument holds for all three angles in the triangle, and so: :$\dfrac c {\sin C} = \dfrac b {\sin B} = \dfrac a {\sin A} = 2 R$ {{qed}}
{{begin-eqn}} {{eqn | l = \sin b \sin c \cos A       | r = \cos a - \cos b \cos c       | c = Spherical Law of Cosines }} {{eqn | ll= \leadsto       | l = \sin^2 b \sin^2 c \cos^2 A       | r = \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c       | c =  }} {{eqn | ll= \leadsto       | l = \sin^2 b \sin^2 c \paren {1 - \sin^2 A}       | r = \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c       | c = Sum of Squares of Sine and Cosine }} {{eqn | ll= \leadsto       | l = \sin^2 b \sin^2 c - \sin^2 b \sin^2 c \sin^2 A       | r = \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c       | c = multiplying out }} {{eqn | ll= \leadsto       | l = \paren {1 - \cos^2 b} \paren {1 - \cos^2 c} - \sin^2 b \sin^2 c \sin^2 A       | r = \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c       | c = Sum of Squares of Sine and Cosine }} {{eqn | ll= \leadsto       | l = 1 - \cos^2 b - \cos^2 c + \cos^2 b \cos^2 c - \sin^2 b \sin^2 c \sin^2 A       | r = \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c       | c = multiplying out }} {{eqn | n = 1       | ll= \leadsto       | l = \sin^2 b \sin^2 c \sin^2 A       | r = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c       | c = rearranging and simplifying }} {{end-eqn}} Let $X \in \R_{>0}$ such that: :$X^2 \sin^2 a \sin^2 b \sin^2 c = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$ Then from $(1)$: {{begin-eqn}} {{eqn | l = \dfrac {X^2 \sin^2 a \sin^2 b \sin^2 c} {\sin^2 b \sin^2 c \sin^2 A}       | o = =       | r = \dfrac {1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c} {1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c}       | c =  }} {{eqn | ll= \leadsto       | l = X^2       | r = \dfrac {\sin^2 A} {\sin^2 a}       | c =  }} {{end-eqn}} In a spherical triangle, all of the sides are less than $\pi$ radians. The same applies to the angles. From Shape of Sine Function: :$\sin \theta > 0$ for all $0 < \theta < \pi$ Hence the negative root of $\dfrac {\sin^2 A} {\sin^2 a}$ does not apply, and so: :$X = \dfrac {\sin A} {\sin a}$ Similarly, from applying the Spherical Law of Cosines to $\cos B$ and $\cos C$: {{begin-eqn}} {{eqn | l = \sin a \sin c \cos B       | r = \cos b - \cos a \cos c }} {{eqn | l = \sin a \sin b \cos C       | r = \cos c - \cos a \cos b }} {{end-eqn}} we arrive at the same point: {{begin-eqn}} {{eqn | l = X       | r = \dfrac {\sin B} {\sin b} }} {{eqn | r = \dfrac {\sin A} {\sin a} }} {{end-eqn}} where: :$X^2 \sin^2 a \sin^2 b \sin^2 c = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$ as before. Hence we have: :$\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$ {{qed}}
:500px Let $A$, $B$ and $C$ be the vertices of a spherical triangle on the surface of a sphere $S$. By definition of a spherical triangle, $AB$, $BC$ and $AC$ are arcs of great circles on $S$. By definition of a great circle, the center of each of these great circles is $O$. Let $O$ be joined to each of $A$, $B$ and $C$. Let $P$ be an arbitrary point on $OC$. Construct $PQ$ perpendicular to $OA$ meeting $OA$ at $Q$. Construct $PR$ perpendicular to $OB$ meeting $OB$ at $R$. In the plane $OAB$: :construct $QS$ perpendicular to $OA$ :construct $RS$ perpendicular to $OB$ where $S$ is the point where $QS$ and $RS$ intersect. Let $OS$ and $PS$ be joined. Let tangents be constructed at $A$ to the arcs of the great circles $AC$ and $AB$. These tangents contain the spherical angle $A$. But by construction, $QS$ and $QP$ are parallel to these tangents Hence $\angle PQS = \sphericalangle A$. Similarly, $\angle PRS = \sphericalangle B$. Also we have: {{begin-eqn}} {{eqn | l = \angle COB       | r = a }} {{eqn | l = \angle COA       | r = b }} {{eqn | l = \angle AOB       | r = c }} {{end-eqn}} It is to be proved that $PS$ is perpendicular to the plane $AOB$. By construction, $OQ$ is perpendicular to both $PQ$ and $QS$. Thus $OQ$ is perpendicular to the plane $PQS$. Similarly, $OR$ is perpendicular to the plane $PRS$. Thus $PS$ is perpendicular to both $OQ$ and $OR$. Thus $PS$ is perpendicular to every line in the plane of $OQ$ and $OR$. That is, $PS$ is perpendicular to the plane $OAB$. In particular, $PS$ is perpendicular to $OS$, $SQ$ and $SR$ It follows that $\triangle PQS$ and $\triangle PRS$ are right triangles. From the right triangles $\triangle OQP$ and $\triangle ORP$, we have: {{begin-eqn}} {{eqn | n = 1       | l = PQ       | r = OP \sin b }} {{eqn | n = 2       | l = PR       | r = OP \sin a }} {{eqn | n = 3       | l = OQ       | r = OP \cos b }} {{eqn | n = 4       | l = OR       | r = OP \cos a }} {{end-eqn}} From the right triangles $\triangle PQS$ and $\triangle PRS$, we have: {{begin-eqn}} {{eqn | l = PS       | r = PS \sin \angle PRS }} {{eqn | r = PQ \sin A }} {{eqn | l = PS       | r = PR \sin \angle PRS }} {{eqn | r = PR \sin B }} {{eqn | ll= \leadsto       | l = OP \sin b \sin A       | r = OP \sin a \sin B       | c = from $(1)$ and $(2)$ }} {{eqn | ll= \leadsto       | l = \dfrac {\sin a} {\sin A}       | r = \dfrac {\sin b} {\sin B}       | c =  }} {{end-eqn}} The result follows by applying this technique mutatis mutandis to the other angles of $ABC$. {{qed}}
Let $B = \sup S$. Let $T = \set {x \in \R: -x \in S}$. Since $\forall x \in S: x \le B$ it follows that $\forall x \in S: -x \ge -B$. Hence $-B$ is a lower bound for $T$. Thus $\set {x \in \R: -x \in S}$ is bounded below. If $C$ is the infimum of $T$, it follows that $C \ge -B$. On the other hand: :$\forall y \in T: y \ge C$ Therefore: :$\forall y \in T: -y \le -C$ Since $S = \set {x \in \R: -x \in T}$ it follows that $-C$ is an upper bound for $S$. Therefore $-C \ge B$ and so $C \le -B$. So $C \le -B$ and $C \ge -B$ and the result follows. {{qed}}
As $T$ is non-empty and bounded below, it follows by the Continuum Property that $T$ admits an infimum. Let $B = \inf T$. Let $S = \set {x \in \R: -x \in T}$. Since $\forall x \in T: x \ge B$ it follows that: :$\forall x \in T: -x \le -B$ Hence $-B$ is an upper bound for $S$, and so $\set {x \in \R: -x \in T}$ is bounded above. If $C$ is the supremum of $S$, it follows that $C \le -B$. On the other hand: :$\forall y \in S: y \le C$ Therefore: :$\forall y \in S: -y \ge -C$ Since $T = \set {x \in \R: -x \in S}$ it follows that $-C$ is a lower bound for $T$. Therefore $-C \le B$ and so $C \ge -B$. So $C \ge -B$ and $C \le -B$ and the result follows. {{qed}}
From Negative of Infimum is Supremum of Negatives, we have that: :$\displaystyle -\inf_{x \mathop \in T} x = \map {\sup_{x \mathop \in T} } {-x} \implies \inf_{x \mathop \in T} x = -\map {\sup_{x \mathop \in T} } {-x}$ Let $S = \set {x \in \R: -x \in T}$. From Negative of Infimum is Supremum of Negatives, $S$ is bounded above. We have: {{begin-eqn}} {{eqn | l = \map {\sup_{x \mathop \in S} } {x + \xi}       | r = \xi + \map {\sup_{x \mathop \in S} } x       | c = Supremum Plus Constant }} {{eqn | ll= \leadsto       | l = \map {\inf_{x \mathop \in T} } {x + \xi}       | r = -\map {\sup_{x \mathop \in T} } {-x + \xi}       | c =  }} {{eqn | r = \xi - \map {\sup_{x \mathop \in T} } {-x}       | c =  }} {{eqn | r = \xi + \map {\inf_{x \mathop \in T} } x       | c =  }} {{end-eqn}} {{qed}}
Let $B = \map \inf S$. Then $B$ is a lower bound for $S$. As $T \subseteq S$, it follows by the definition of a subset that $x \in T \implies x \in S$. Because $x \in S \implies B \preceq x$ (as $B$ is a lower bound for $S$) it follows that $x \in T \implies B \preceq x$. So $B$ is a lower bound for $T$. Therefore $B$ precedes the infimum of $T$ in $U$. Hence the result. {{qed}}
As $A$ is the center of circle $BCD$, it follows from {{EuclidDefLink|I|15|Circle}} that $AC = AB$. As $B$ is the center of circle $ACE$, it follows from {{EuclidDefLink|I|15|Circle}} that $BC = AB$. So, as $AC = AB$ and $BC = AB$, it follows from {{EuclidCommonNotionLink|1}} that $AC = BC$. Therefore $AB = AC = BC$. Therefore $\triangle ABC$ is equilateral. {{qed}} {{Euclid Note|1|I}}
As $B$ is the center of circle $CGH$, it follows from {{EuclidDefLink|I|15|Circle}} that $BC = BG$. As $D$ is the center of circle $GKL$, it follows from {{EuclidDefLink|I|15|Circle}} that $DL = DG$. As $\triangle ABD$ is an equilateral triangle, it follows that $DA = DB$. Therefore, by Common Notion 3, $AL = BG$. As $AL = BG$ and $BC = BG$, it follows from Common Notion 1 that $AL = BC$. Therefore, at the given point $A$, the required straight line segment $AL$ has been placed equal in length to $BC$. {{qed}} {{Euclid Note|2|I}}
As $A$ is the center of circle $DEF$, it follows from {{EuclidDefLink|I|15|Circle}} that $AE = AD$. But $C$ is also equal to $AD$. So, as $C = AD$ and $AD = AE$, it follows from Common Notion 1 that $AE = C$. Therefore, given the two straight line segments $AB$ and $C$, from the greater of these $AB$, a length $AE$ has been cut off equal to the lesser $C$. {{qed}} {{Euclid Note|3|I}}
500px Let $\triangle ABC$ and $\triangle DEF$ be $2$ triangles having sides $AB = DE$ and $AC = DF$, and with $\angle BAC = \angle EDF$. If $\triangle ABC$ is placed on $\triangle DEF$ such that: : the point $A$ is placed on point $D$, and  : the line $AB$ is placed on line $DE$ then the point $B$ will also coincide with point $E$ because $AB = DE$. So, with $AB$ coinciding with $DE$, the line $AC$ will coincide with the line $DF$ because $\angle BAC = \angle EDF$. Hence the point $C$ will also coincide with the point $F$, because $AC = DF$. But $B$ also coincided with $E$. Hence the line $BC$ will coincide with line $EF$. (Otherwise, when $B$ coincides with $E$ and $C$ with $F$, the line $BC$ will not coincide with line $EF$ and two straight lines will enclose a region which is impossible.) Therefore $BC$ will coincide with $EF$ and be equal to it. Thus the whole $\triangle ABC$ will coincide with the whole $\triangle DEF$ and thus $\triangle ABC = \triangle DEF$. The remaining angles on $\triangle ABC$ will coincide with the remaining angles on $\triangle DEF$ and be equal to them. {{qed}} {{Euclid Note|4|I}}
200px Let $\triangle ABC$ be an isosceles triangle whose side $AB$ equals side $AC$. We extend the straight lines $AB$ and $AC$ to $D$ and $E$ respectively. Let $F$ be a point on $BD$. We cut off from $AE$ a length $AG$ equal to $AF$. We draw line segments $FC$ and $GB$. Since $AF = AG$ and $AB = AC$, the two sides $FA$ and $AC$ are equal to $GA$ and $AB$ respectively. They contain a common angle, that is, $\angle FAG$. So by Triangle Side-Angle-Side Equality, $\triangle AFC = \triangle AGB$. Thus $FC = GB$, $\angle ACF = \angle ABG$ and $\angle AFC = \angle AGB$. Since $AF = AG$ and $AB = AC$, then $BF = CG$. But $FC = GB$, so the two sides $BF, FC$ are equal to the two sides $CG, GB$ respectively. Then $\angle BFC = \angle CGB$ while $CB$ is common to both. Therefore by Triangle Side-Angle-Side Equality, $\triangle BFC = \triangle CGB$. Therefore $\angle FBC = \angle GCB$ and $\angle BCF = \angle CBG$. So since $\angle ACF = \angle ABG$, and in these $\angle BCF = \angle CBG$, then $\angle ABC = \angle ACB$. But $\angle ABC$ and $\angle ACB$ are at the base of $\triangle ABC$. Also, we have already proved that $\angle FBC = \angle GCB$, and these are the angles under the base of $\triangle ABC$. Hence the result. {{qed}}
:200px Let $\triangle ABC$ be a triangle in which $\angle ABC = \angle ACB$. Suppose side $AB$ is not equal to side $AC$. Then one of them will be greater. {{WLOG}}, Suppose $AB > AC$. We cut off from $AB$ a length $DB$ equal to $AC$. We draw the line segment $CD$. Since $DB = AC$, and $BC$ is common, the two sides $DB, BC$ are equal to $AC, CB$ respectively. Also, $\angle DBC = \angle ACB$. So by Triangle Side-Angle-Side Equality, $\triangle DBC = \triangle ACB$. But $\triangle DBC$ is smaller than $\triangle ACB$, which is absurd. Therefore, have $AB \le AC$.  A similar argument shows the converse, and hence $AB = AC$. {{qed}}
Let $\angle ABC$ and $\angle ACB$ be the angles that are the same. {{begin-eqn}} {{eqn | n = 1       | l = \angle ABC       | r = \angle ACB       | c = Given }} {{eqn | n = 2       | l = BC       | r = CB       | c = Equality is Reflexive }} {{eqn | n = 3       | l = \angle ACB       | r = \angle ABC       | c = Given }} {{eqn | n = 4       | l = \triangle ABC       | r = \triangle ACB       | c = Triangle Angle-Side-Angle Equality by $\left({1}\right)$, $\left({2}\right)$ and $\left({3}\right)$ }} {{eqn | n = 5       | l = AB       | r = AC       | c = from $\left({4}\right)$ }} {{end-eqn}} {{qed}}
:400px Let $AC$ and $CB$ be constructed on $AB$ meeting at $C$. Let two other straight line segments $AD$ and $DB$ be constructed on $AB$, on the same side of it, meeting at $D$, such that $AC = AD$ and $CB = DB$. Suppose, with a view to obtaining a contradiction, $C$ and $D$ are different points. Let $CD$ be joined. Since $AC = AD$ it follows that $\angle ACD = \angle ADC$. Therefore $\angle ACD$ is greater than $\angle DCB$ because the whole is greater than the part. Therefore $\angle CDB$ is much greater than $\angle DCB$. Now since $CB = DB$, it follows that $\angle CDB = \angle DCB$. But it was proved much greater than it. From this contradiction it follows that $C$ and $D$ can not be different points. Hence the result. {{qed}} {{Euclid Note|7|I}}
:500px Let $\triangle ABC$ and $\triangle DEF$ be two triangles such that: : $AB = DE$ : $AC = DF$ : $BC = EF$ Suppose $\triangle ABC$ were superimposed over $\triangle DEF$ so that point $B$ is placed on point $E$ and the side $BC$ on $EF$. Then $C$ will coincide with $F$, as $BC = EF$ and so $BC$ coincides with $EF$. {{AimForCont}} $BA$ does not coincide with $ED$ and $AC$ does not coincide with $DF$. Then they will fall as, for example, $EG$ and $GF$. Thus there will be two pairs of straight line segments constructed on the same line segment, on the same side as it, meeting at different points. This contradicts the theorem Two Lines Meet at Unique Point. Therefore $BA$ coincides with $ED$ and $AC$ coincides with $DF$. Therefore $\angle BAC$ coincides with $\angle EDF$ and is equal to it. The same argument can be applied to the other two sides, and thus we show that all corresponding angles are equal. {{qed}} {{Euclid Note|8|I}}
We have: : $AD = AE$ : $AF$ is common : $DF = EF$ Thus triangles $\triangle ADF$ and $\triangle AEF$ are equal. Thus $\angle DAF = \angle EAF$. Hence $\angle BAC$ has been bisected by $AF$. {{qed}} {{Euclid Note|9|I|There are quicker and easier constructions of a bisection, but this particular one uses only results previously demonstrated.}}
As $\triangle ABC$ is an equilateral triangle, it follows that $AC = CB$. The two triangles $\triangle ACD$ and $\triangle BCD$ have side $CD$ in common, and side $AC$ of $\triangle ACD$ equals side $BC$ of $\triangle BCD$. The angle $\angle ACD$ subtended by lines $AC$ and $CD$ equals the angle $\angle BCD$ subtended by lines $BC$ and $CD$, as $\angle ACB$ was bisected. So triangles $\triangle ACD$ and $\triangle BCD$ are equal. Therefore $AD = DB$. So $AB$ has been bisected at the point $D$. {{qed}} {{Euclid Note|10|I}}
Since $DC = CE$ and $FC$ is common to both, and $DF = FE$, triangle $\triangle DCF$ equals triangle $\triangle ECF$. Thus $\angle DCF = \angle ECF$. So $CF$ is a straight line set up on a straight line making the adjacent angles equal to one another. Thus it follows from {{EuclidDefLink|I|10|Right Angle}} that each of $\angle DCF$ and $\angle ECF$ are right angles. So the straight line $CF$ has been drawn at right angles to the given straight line $AB$ from the given point $C$ on it. {{qed}} {{Euclid Note|11|I}}
As $C$ is the center of circle $BCD$, it follows from {{EuclidDefLink|I|15|Circle}} that $GC = CE$. As $EG$ has been bisected, $GH = HE$. Thus, as $GC = CE$ and $GH = HE$, and $CH$ is common, by Triangle Side-Side-Side Equality, $\triangle CHG = \triangle EHG$. Therefore $\angle CHG = \angle CHE$. So $CH$ is a straight line set up on a straight line making the adjacent angles equal to one another. Thus it follows from {{EuclidDefLink|I|10|Right Angle}} that each of $\angle CHG$ and $\angle CHE$ are right angles. So the straight line $CH$ has been drawn at right angles to the given infinite straight line $AB$ through the given point $C$. {{qed}} {{Euclid Note|12|I|This theorem is one of the two attributed to {{AuthorRef|Oenopides of Chios}} (the other being {{EuclidPropLink|prop = 23|book = I|title = Construction of Equal Angle}}).}}
:400px Let the line $AB$ be set up on the line $CD$. If $\angle ABC = \angle ABD$ then from {{EuclidDefLink|I|10|Right Angle}} they are two right angles. If not, we draw $BE$ at right angles to $CD$. Then: :$\angle CBE = \angle EBD$ which from {{EuclidDefLink|I|10|Right Angle}} are both right angles.  Since $\angle CBE = \angle ABC + \angle ABE$, we add $\angle EBD$ to each. Therefore by {{EuclidCommonNotionLink|2}}: :$\angle CBE + \angle EBD = \angle ABC + \angle ABE + \angle EBD$ Therefore: :$\angle ABC + \angle ABD = \angle ABC + \angle ABE + \angle EBD$ Again, since $\angle ABD = \angle EBD + \angle ABE$, we add $\angle ABC$ to each. Therefore by {{EuclidCommonNotionLink|2}}: :$\angle ABD + \angle ABC = \angle ABC + \angle ABE + \angle EBD$ But $\angle CBE + \angle EBD$ equals the same three angles. Therefore by {{EuclidCommonNotionLink|1}}: :$\angle CBE + \angle EBD = \angle ABC + \angle ABD$ But $\angle CBE$ and $\angle EBD$ are both right angles. Hence the result. {{qed}} {{Euclid Note|13|I|{{EuclidNoteConverse|prop = 14|title = Two Angles making Two Right Angles make Straight Line}}}}
400px Let $AB$ be the given straight line and $B$ the given point on it. Let $BC$ and $BD$ be the two straight lines not lying on the same side of it such that $\angle ABC + \angle ABD$ equal two right angles. Suppose $BD$ is not in a straight line with $BC$. Now some line must be, so let $BE$ be in a straight line with $BC$ instead. Then since $AB$ stands on the (supposed) straight line $CBE$, $\angle ABC + \angle ABE$ equal two right angles. But we already know that $\angle ABC + \angle ABD$ equal two right angles. Therefore, by Common Notion 1 and the fact that all right angles are congruent, $\angle ABC + \angle ABD = \angle ABC + \angle ABE$. So let $\angle ABC$ be subtracted from each. Then by Common Notion 3, the remaining angle $\angle ABE$ equals remaining angle $\angle ABD$. But $\angle ABD$ is greater than $\angle ABE$, so this is impossible. So $BE$ can not be in a straight line with $BC$. Similarly we can show that ''any'' straight line which is not $BD$ can not be in a straight line with $BC$. Therefore $BC$ is in a straight line with $BD$, hence the result. {{qed}} {{Euclid Note|14|I|{{EuclidNoteConverse|prop = 13|title = Two Angles on Straight Line make Two Right Angles}}}}
300px Let $AB$ and $CD$ be two straight lines that cut each other at the point $E$. Since the straight line $AE$ stands on the straight line $CD$, the angles $\angle AED$ and $\angle AEC$ make two right angles. Since the straight line $DE$ stands on the straight line $AB$, the angles $\angle AED$ and $\angle BED$ make two right angles. But $\angle AED$ and $\angle AEC$ also make two right angles. So by Common Notion 1 and the fact that all right angles are congruent, $\angle AED + \angle AEC = \angle AED + \angle BED$. Let $\angle AED$ be subtracted from each. Then by Common Notion 3 it follows that $\angle AEC = \angle BED$. Similarly it can be shown that $\angle BEC = \angle AED$. {{qed}}
:250px Let $\triangle ABC$ be a triangle. Let the side $BC$ be extended to $D$. Let $AC$ be bisected at $E$. Let $BE$ be joined and extended to $F$. Let $EF$ be made equal to $BE$. (Technically we really need to extend $BE$ to a point beyond $F$ and then crimp off a length $EF$.) Let $CF$ be joined. Let $AC$ be extended to $G$. We have $\angle AEB = \angle CEF$ from Two Straight Lines make Equal Opposite Angles. Since $AE = EC$ and $BE = EF$, from Triangle Side-Angle-Side Equality we have $\triangle ABE = \triangle CFE$. Thus $AB = CF$ and $\angle BAE = \angle ECF$. But $\angle ECD$ is greater than $\angle ECF$. Therefore $\angle ACD$ is greater than $\angle BAE$. Similarly, if $BC$ were bisected, $\angle BCG$, which is equal to $\angle ACD$ by Two Straight Lines make Equal Opposite Angles, would be shown to be greater than $\angle ABC$ as well. Hence the result. {{qed}} {{Euclid Note|16|I}}
:300px Let $\triangle ABC$ be a triangle. Let the side $BC$ be extended to $D$. Since the angle $\angle ACD$ is an external angle of $\triangle ABC$, it follows that it is greater than both $\angle BAC$ and $\angle ABC$. We add $\angle ACB$ to both, so that $\angle ACD + \angle ACB$ is greater than $\angle ABC + \angle ACB$. But $\angle ACD + \angle ACB$ is equal to two right angles. Therefore $\angle ABC + \angle ACB$ is less than two right angles. In a similar manner we show that the same applies to the other two pairs of internal angles of $\triangle ABC$. {{qed}} {{Euclid Note|17|I}}
:350px Let $\triangle ABC$ be a triangle such that $AC$ is greater than $AB$. Let $AD$ be made equal to $AB$. Let $BD$ be joined. Then $\angle ADB$ is an exterior angle of the triangle $\triangle BCD$. Therefore $\angle ADB$ is greater than $\angle ACB$. As $AD = AB$, the triangle $\triangle ABD$ is isosceles. From Isosceles Triangle has Two Equal Angles, $\angle ADB = \angle ABD$. Therefore $\angle ABD$ is greater than $\angle ACB$. Therefore, as $\angle ABC = \angle ABD + \angle DBC$, it follows that $\angle ABC$ is greater than $\angle ACB$. Hence the result. {{qed}} {{Euclid Note|18|I|{{EuclidNoteConverse|prop = 19|title = Greater Angle of Triangle Subtended by Greater Side}}}}
:200px Let $\triangle ABC$ be a triangle such that $\angle ABC$ is greater than $\angle BCA$. Suppose $AC$ is not greater than $AB$. If $AC$ were equal to $AB$, then by Isosceles Triangle has Two Equal Angles, $\angle ABC = \angle BCA$, but they're not so it isn't. If $AC$ were less than $AB$, then by Greater Side of Triangle Subtends Greater Angle it would follow that $\angle ABC$ is less than $\angle BCA$, but it's not so it isn't. So $AC$ must be greater than $AB$ Hence the result. {{qed}} {{Euclid Note|19|I|{{EuclidNoteConverse|prop = 18|title = Greater Side of Triangle Subtends Greater Angle}}}}