hcju/proofwikiqa huggingface dataset:
First we show that: :$\displaystyle \limsup_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\sup_{k \mathop \ge n} x_k}$ Let $M_n = \displaystyle \sup_{k \mathop \ge n} x_k$. By Supremum of Subset, the sequence $\sequence {M_n}$ decreases, for: :$m \ge n \implies \set {k \in \N: k \ge m} \subseteq \set {k \in \N: k \ge n}$ A lower bound for $\sequence {x_n}$ is also a lower bound for $\sequence {M_n}$. So from the Monotone Convergence Theorem (Real Analysis) it follows that $\sequence {M_n}$ converges. Suppose $M_n \to M$ as $n \to \infty$. From the Bolzano-Weierstrass Theorem there exists a convergent subsequence of $\sequence {x_n}$. Let $L$ be the set of all numbers which are the limit of some subsequence of $\sequence {x_n}$. Let $l \in L$ be arbitrary. Let $\sequence {x_{n_r} }$ be a convergent subsequence of $\sequence {x_n}$ such that $x_{n_r} \to l$ as $r \to \infty$. Then $\forall n_r \ge n: x_{n_r} \le M_n$. Hence $l \le M_n$ by Lower and Upper Bounds for Sequences and hence (from the same theorem) $l \le M$. This is true for all $l \in L$, so $\overline l = \limsup_{n \mathop \to \infty} x_n \le M$. Now, from Terms of Bounded Sequence Within Bounds, we have that: :$\forall \epsilon > 0: \exists n: \forall k \ge n: x_k < \overline l + \epsilon$ Thus $\overline l + \epsilon$ is an upper bound for $\set {x_k: k \ge n}$. So $M \le M_n \le \overline l + \epsilon$. Thus from Real Plus Epsilon, $M \le \overline l$. Thus we conclude that $M = \overline l$ and hence: :$\displaystyle \overline l = \limsup_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\sup_{k \ge n} x_k}$ {{handwaving|"similar argument"}} :$\displaystyle \liminf_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop to \infty} } {\inf_{k \mathop \ge n} x_k}$ can be proved using a similar argument. {{qed}}
=== Absolute Convergence === Let $l < 1$. Then let us choose $\epsilon > 0$ such that $l + \epsilon < 1$. Consider the real sequence $\sequence {b_n}$ defined by $\sequence {b_n} = \sequence {\size {a_n} }$. Here, $\size {a_n}$ denotes either the absolute value of $a_n$, or the complex modulus of $a_n$. Then: :$\displaystyle l = \limsup_{n \mathop \to \infty} {b_n}^{1/n}$ It follows from Terms of Bounded Sequence Within Bounds that for sufficiently large $n$,: :$b_n < \paren {l + \epsilon}^n$ By Sum of Infinite Geometric Sequence, the series $\displaystyle \sum_{n \mathop = 1}^\infty \paren {l + \epsilon}^n$ converges. By the comparison test, $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ converges. Hence $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges absolutely by the definition of absolute convergence. {{qed|lemma}} === Divergence === Let $l > 1$. Then we choose $\epsilon > 0$ such that $l - \epsilon > 1$. {{AimForCont}} that there exist an upper bound for the set: :$S := \set {n \in \N: \size {a_n}^{1/n} > l - \epsilon}$ Then for all sufficiently large $n$: :$\size {a_n}^{1/n} \le l - \epsilon$ However, this implies that: :$\displaystyle \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} \le l - \epsilon$ which is false by the definition of $l$. The set $S$, then, is not bounded. This means that there exist arbitrarily large $n$ such that: :$\size {a_n} > \paren {l - \epsilon}^n$ Thus: :$\displaystyle \lim_{n \mathop \to \infty} \size {a_n} \ne 0$ and so $\displaystyle \lim_{n \mathop \to \infty} a_n \ne 0$. Hence from Terms in Convergent Series Converge to Zero, $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ must be divergent. {{qed}}
If $x = 0$ the result is trivially true as: :$\forall n \ge 1: \dfrac {0^n} {n!} = 0$ If $x \ne 0$ we have: :$\left|{\dfrac{\left({\dfrac {x^{n+1}} {(n+1)!}}\right)}{\left({\dfrac {x^n}{n!}}\right)}}\right| = \dfrac {\left|{x}\right|} {n+1} \to 0$ as $n \to \infty$. This follows from the results: :Sequence of Powers of Reciprocals is Null Sequence, where $\dfrac 1 n \to 0$ as $n \to \infty$ :The Squeeze Theorem for Real Sequences, as $\dfrac 1 {n + 1} < \dfrac 1 n$ :The Multiple Rule for Real Sequences, putting $\lambda = \left|{x}\right|$. Hence by the Ratio Test: $\displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ converges. {{qed}} Alternatively, the Comparison Test could be used but this is more cumbersome in this instance. Another alternative is to view this as an example of Radius of Convergence of Power Series over Factorial setting $\xi = 0$.
:400px Construct a point $D$ so that $\Box ABDC$ is a parallelogram. From Halves of Parallelogram Are Congruent Triangles: :$\triangle ABC \cong \triangle DCB$ hence their areas are equal. The Area of Parallelogram is equal to the product of one of its bases and the associated altitude. Thus  {{begin-eqn}} {{eqn | l = \paren {ABCD}       | r = c \cdot h_c }} {{eqn | ll= \leadsto       | l = 2 \paren {ABC}       | r = c \cdot h_c       | c = because congruent surfaces have equal areas }} {{eqn | l = \paren {ABC}       | r = \frac {c \cdot h_c} 2 }} {{end-eqn}} where $\paren {XYZ}$ is the area of the plane figure $XYZ$. A similar argument can be used to show that the statement holds for the other sides. {{qed}}
That $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent means that $\displaystyle \sum_{n \mathop = 1}^\infty \norm {a_n}$ converges in $\R$. Hence the sequence of partial sums is a Cauchy sequence by Convergent Sequence is Cauchy Sequence. Now let $\epsilon > 0$. Let $N \in \N$ such that for all $m, n \in \N$, $m \ge n \ge N$ implies that: :$\displaystyle \sum_{k \mathop = n + 1}^m \norm {a_k} = \size {\sum_{k \mathop = 1}^m \norm {a_k} - \sum_{k \mathop = 1}^n \norm {a_k} } < \epsilon$ This $N$ exists because the sequence is Cauchy. Now observe that, for $m \ge n \ge N$, one also has: {{begin-eqn}} {{eqn | l = \norm {\sum_{k \mathop = 1}^m a_k - \sum_{k \mathop = 1}^n a_k}       | r = \norm {\sum_{k \mathop = n + 1}^m a_k} }} {{eqn | o = \le       | r = \sum_{k \mathop = n + 1}^m \norm {a_k}       | c = Triangle inequality for $\norm {\, \cdot \,}$ }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} It follows that the sequence of partial sums of $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is Cauchy. As $V$ is a Banach space, this implies that $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges. {{qed}}
Let $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ be convergent. Then by Terms in Convergent Series Converge to Zero, $\left \langle {b_n} \right \rangle$ converges to zero. A Convergent Sequence is Bounded. So it follows that: : $\exists H: \forall n \in \N_{>0}: a_n \le H b_n$ Thus, by the corollary to the Comparison Test, $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is convergent. Since $l > 0$, from Sequence Converges to Within Half Limit: : $\exists N: \forall n > N: a_n > \dfrac 1 2 l b_n$ Hence the convergence of $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ implies the convergence of $\displaystyle \sum_{n \mathop = 1}^\infty b_n$. {{qed}}
There are three cases to be analysed: the square, the rectangle and the general parallelogram. === Square === {{:Area of Parallelogram/Square}} === Rectangle === {{:Area of Parallelogram/Rectangle}} === Parallelogram === {{:Area of Parallelogram/Parallelogram}}
=== Proof of the First Part === First, we show that the area is equal to $\rho_a \left({s - a}\right) = \rho_b \left({s - b}\right) = \rho_c \left({s - c}\right)$. {{WLOG}}, we pick an excircle $I_a$. File:Area1.PNG {{begin-eqn}} {{eqn | l = (ABC)       | r = (ABI_a) + (ACI_a) - (CBI_a)       | c = (see figure above) }} {{eqn | r = \frac {c \rho_a} 2 + \frac {b \rho_a} 2 - \frac {a \rho_a} 2       | c = Area of Triangle in Terms of Side and Altitude }} {{eqn | r = \rho_a \frac {b + c + a} 2 - \rho_a a }} {{eqn | r = \rho_a s - \rho_a a }} {{eqn | l = (ABC)       | r = \rho_a \left({s - a}\right) }} {{end-eqn}} A similar argument can be used to show that the statement holds for the other excircles. {{qed|lemma}} === Proof of the Second Part === Second, we show the area is equal to $(ABC) = \rho s$. We take the incircle with incenter at $I$ and inradius $\rho$: File:T3.PNG {{begin-eqn}} {{eqn | l = (ABC)       | r = (ABI) + (BCI) + (CAI)       | c = (see figure above) }} {{eqn | r = \frac {a \rho} 2 + \frac {b \rho} 2 + \frac {c \rho} 2 }} {{eqn | r = \frac {\rho} 2 \left({a + b + c}\right) }} {{eqn | l = (ABC)       | r = \rho s }} {{end-eqn}} {{qed|lemma}} === Proof of the Third Part === Finally, we show that the area is equal to $\sqrt{\rho_a \rho_b \rho_c \rho}$: {{begin-eqn}} {{eqn | l = (ABC)^4       | r = \rho_a \left({s - a}\right) \rho_b \left({s - b}\right) \rho_c \left({s - c}\right) \rho s }} {{eqn | l = (ABC)^4       | r = s \left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \rho_a \rho_b \rho_c \rho }} {{eqn | l = (ABC)^4       | r = (ABC)^2 \rho_a \rho_b \rho_c \rho       | c = Heron's Formula }} {{eqn | l = (ABC)^2       | r = \rho_a \rho_b \rho_c \rho }} {{eqn | l = (ABC)       | r = \sqrt{\rho_a \rho_b \rho_c \rho} }} {{end-eqn}} {{Qed}}
:400px Let $O$ be the circumcenter of $\triangle ABC$. Let $\AA$ be the area of $\triangle ABC$. Let a perpendicular be dropped from $C$ to $AB$ at $E$. Let $h := CE$. Then: {{begin-eqn}} {{eqn | l = \AA       | r = \frac {c h} 2       | c = Area of Triangle in Terms of Side and Altitude }} {{eqn | n = 1       | ll= \leadsto       | l = h       | r = \frac {2 \AA} c       | c =  }} {{end-eqn}} Let a diameter $CD$ of the circumcircle be passed through $O$. By definition of circumradius, $CD = 2 R$. By Thales' Theorem, $\angle CAD$ is a right angle. By Angles on Equal Arcs are Equal, $\angle ADC = \angle ABC$. It follows from Sum of Angles of Triangle equals Two Right Angles that $\angle ACD = \angle ECB$. Thus by Equiangular Triangles are Similar $\triangle DAC$ and $\triangle BEC$ are similar. So: {{begin-eqn}} {{eqn | l = \frac {CA} {CD}       | r = \frac {CE} {CB}       | c = $\triangle DAC$ and $\triangle BEC$ are similar }} {{eqn | ll= \leadsto       | l = \frac b {2 R}       | r = \frac h a       | c =  }} {{eqn | r = \frac {2 \AA} {a c}       | c = substituting for $h$ from $(1)$ above }} {{eqn | ll= \leadsto       | l = \AA       | r = \frac {a b c} {4 R}       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | n = 1       | l = b^2       | r = AP^2 + CP^2 - 2 AP \cdot CP \cdot \map \cos {\angle APC}       | c = Law of Cosines }} {{eqn | n = 2       | l = a^2       | r = PB^2 + CP^2 - 2 CP \cdot PB \cdot \map \cos {\angle BPC}       | c = Law of Cosines }} {{eqn | r = PB^2 + CP^2 + 2 CP \cdot PB \cdot \map \cos {\angle APC}       | c = Cosine of Supplementary Angle }} {{eqn | n = 3       | ll= \leadsto \quad       | l = b^2 \cdot PB       | r = AP^2 \cdot PB + CP^2 \cdot PB - 2 PB \cdot AP \cdot CP \cdot \map \cos {\angle APC}       | c = $(1) \ \times PB$ }} {{eqn | n = 4       | l = a^2 \cdot AP       | r = PB^2 \cdot AP + CP^2 \cdot AP + 2 AP \cdot CP \cdot PB \cdot \map \cos {\angle APC}       | c = $(2) \ \times AP$ }} {{eqn | ll= \leadsto \quad       | l = a^2 \cdot AP + b^2 \cdot PB       | r = AP^2 \cdot PB + PB^2 \cdot AP + CP^2 \cdot PB + CP^2 \cdot AP       | c = $(3) \ + \ (4)$ }} {{eqn | r = CP^2 \left({PB + AP}\right) + AP \cdot PB \paren {PB + AP} }} {{eqn | r = c \paren {CP^2 + AP \cdot PB}       | c = as $PB + AP = c$ }} {{end-eqn}} {{qed}} {{namedfor|Matthew Stewart|cat = Stewart}} It is also known as Apollonius's Theorem after {{AuthorRef|Apollonius of Perga}}. Category:Triangles knuni2ihszsgiqh1ceg11nxve4zy050
Let $\triangle ABC$ be embedded in the complex plane. :300px Let $A = \tuple {x_1, y_1}$, $B = \tuple {x_2, y_2}$ and $C = \tuple {x_3, y_3}$ be represented by the complex numbers $z_1$, $z_2$ and $z_3$ respectively. Then: {{begin-eqn}} {{eqn | l = AC       | r = z_3 - z_1       | c =  }} {{eqn | l = BC       | r = z_3 - z_2       | c =  }} {{eqn | l = AB       | r = z_2 - z_1       | c =  }} {{eqn | l = AD       | r = \dfrac {AB} 2       | c =  }} {{eqn | r = \paren {\dfrac {z_2 - z_1} 2}       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = AC + CD       | r = AD       | c =  }} {{eqn | ll= \leadsto       | l = CD       | r = AD - AC       | c =  }} {{eqn | r = \paren {\dfrac {z_2 - z_1} 2} - \paren {z_3 - z_1}       | c =  }} {{eqn | r = \paren {\dfrac {z_1 - z_3} 2} + \paren {\dfrac {z_2 - z_3} 2}       | c =  }} {{end-eqn}} {{finish}}
=== Definition 1 implies Definition 2 === Let $S$ be a bounded subset of the real numbers by definition 1. Then by definition $S$ is bounded both above and below. As $S$ is bounded below: :$\exists L \in \R: \forall x \in S: L \le x$ As $S$ is bounded above: :$\exists H \in \R: \forall x \in S: H \ge x$ Let: :$K = \max \set {\size L, \size H}$ Then: :$K \ge \size L$ and $K \ge \size H$ It follows from Negative of Absolute Value: Corollary 2 that $-K \le L \le K$ and $-K \le H \le K$. In particular: :$-K \le L$ and so $\forall x \in S: -K \le x$ :$H \le K$ and so $\forall x \in S: x \le K$ Thus: :$\forall x \in S: -K \le x \le K$ and, by Negative of Absolute Value: Corollary 2: :$\forall x \in S: \size x \le K$ Thus $S$ is a bounded subset of the real numbers by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $S$ be a bounded subset of the real numbers by definition 2. Thus: :$\exists K \in \R: \forall x \in S: \size x \le K$ Then by Negative of Absolute Value: Corollary 2: :$\forall x \in S: -K \le x \le K$ Thus by definition, $S$ is both bounded above (by $K$) and bounded below (by $-K$). Thus $S$ is a bounded subset of the real numbers by definition 1. {{qed}}
{{begin-eqn}} {{eqn | l = \frac {BD} {DC}        | r = \frac c b       | c = Angle Bisector Theorem }} {{eqn | ll= \leadsto       | l = \frac {BD} {DC} + 1       | r = \frac c b + 1 }} {{eqn | ll= \leadsto       | l = \frac {BD + DC} {DC}       | r = \frac {b + c} b }} {{eqn | ll= \leadsto       | l = \frac a {DC}       | r = \frac {b + c} b }} {{eqn | ll= \leadsto       | l = DC       | r = \frac {a b} {b + c} }} {{end-eqn}} Similarly, or by symmetry, we get: :$BD = \dfrac {a c} {b + c}$ From Stewart's Theorem, we have: :$b^2 \cdot BD + c^2 \cdot DC = d^2 \cdot a + BD \cdot DC \cdot a$ Substituting the above expressions for $BD$ and $DC$: {{begin-eqn}} {{eqn | l = b^2 \dfrac {a c} {b + c} + c^2 \frac {a b} {b + c}       | r = d^2 \cdot a + \dfrac {a c} {b + c} \cdot \frac {a b} {b + c} \cdot a }} {{eqn | ll= \leadsto       | l = a b c \frac {b + c} {b + c}       | r = d^2 \cdot a + \frac{a^2 b c} {\paren {b + c}^2} \cdot a }} {{eqn | ll= \leadsto       | l = b c       | r = d^2 + \frac {a^2 b c} {\paren {b + c}^2} }} {{eqn | ll= \leadsto       | l = d^2       | r = b c \paren {1 - \frac {a^2} {\paren {b + c}^2} } }} {{eqn | ll= \leadsto       | l = d^2       | r = \frac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2} }} {{end-eqn}} {{qed}}
:400px From Length of Angle Bisector: Proof 1, we have: :$BD = \dfrac {a c} {b + c}$ :$DC = \dfrac {a b} {b + c}$ Then we have: {{begin-eqn}} {{eqn | l = \angle BAD       | o = \cong       | r = \angle FAC       | c = Definition of Angle Bisector }} {{eqn | l = \angle ABD       | o = \cong       | r = \angle AFC       | c = Angles in Same Segment of Circle are Equal }} {{end-eqn}} Then from Triangles with Two Equal Angles are Similar we have: :$\triangle ABD \sim \triangle AFC$ So: {{begin-eqn}} {{eqn | l = \frac c d       | r = \frac {AF} b       | c = as $\triangle ABD$ and $\triangle AFC$ are similar }} {{eqn | ll= \leadsto       | l = \frac c d       | r = \frac {d + DF} b }} {{end-eqn}} Now we use the Intersecting Chord Theorem, which gives us $BD \cdot DC = d \cdot DF$. {{begin-eqn}} {{eqn | l = \frac c d       | r = \frac {d + \frac {BD \cdot DC} d} b }} {{eqn | ll= \leadsto       | l = b c       | r = d^2 + BD \cdot DC }} {{eqn | ll= \leadsto       | l = d^2       | r = b c - BD \cdot DC }} {{eqn | ll= \leadsto       | l = d^2       | r = b c - \frac {a c} {b + c} \cdot \frac {a b} {b + c} }} {{eqn | ll= \leadsto       | l = d^2       | r = \frac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2} }} {{end-eqn}} {{qed}}
Let $B = \sup \left({S}\right)$. Then $B$ is an upper bound for $S$. As $T \subseteq S$, it follows by the definition of a subset that $x \in T \implies x \in S$. Because $x \in S \implies x \preceq B$ (as $B$ is an upper bound for $S$) it follows that $x \in T \implies x \preceq B$. So $B$ is an upper bound for $T$. Therefore $B$ succeeds the supremum of $T$ in $S$. Hence the result. {{qed}} Category:Order Theory h6yid7mhhlbjzlqg1wmrmg3ej6mowmg
The number $\sup S$ is an upper bound for $S$. Therefore, $\sup S$ is an upper bound for $T$ as $T$ is a non-empty subset of $S$. Accordingly, $T$ has a supremum by the Continuum Property. The number $\sup S$ is an upper bound for $T$. Therefore, $\sup S$ is greater than or equal to $\sup T$ as $\sup T$ is the least upper bound of $T$. {{qed}}
By the Continuum Property, $T$ admits a supremum. It follows from Supremum of Subset that $\sup T \le \sup S$. {{qed}}
$S$ is bounded above as $S$ has a supremum. Therefore, $T$ is bounded above as $T$ is a subset of $S$. Accordingly, $T$ admits a supremum by the Continuum Property as $T$ is non-empty. We know that $\sup T$ and $\sup S$ exist. Therefore by Suprema of two Real Sets: :$\forall \epsilon \in \R_{>0}: \forall t \in T: \exists s \in S: t < s + \epsilon \iff \sup T \le \sup S$ We have: {{begin-eqn}} {{eqn | l = \forall \epsilon       | o = \in       | r = \R_{>0}: 0 < \epsilon }} {{eqn | ll= \leadsto       | l = \forall \epsilon       | o = \in       | r = \R_{>0}: \forall t \in T: t < t + \epsilon }} {{eqn | ll= \leadsto       | l = \forall \epsilon       | o = \in       | r = \R_{>0}: \forall t \in T: t < s + \epsilon \land s = t }} {{eqn | ll= \leadsto       | l = \forall \epsilon       | o = \in       | r = \R_{>0}: \forall t \in T: \exists s \in S: t < s + \epsilon       | c = as $T \subseteq S$ }} {{eqn | ll= \leadsto       | l = \sup T       | o = \le       | r = \sup S }} {{end-eqn}} {{qed}}
By definition $\sup S$ is an upper bound for $S$. Thus: :$\forall x \in S: x \le \sup S$ As $T \subseteq S$ we have by definition of subset that: :$\forall x \in T: x \in S$ Hence: :$\forall x \in T: x \le \sup S$ So by definition $\sup S$ is an upper bound for $T$. So $\sup S$ is at least as big as the smallest upper bound for $T$ Thus by definition of supremum: :$\sup T \le \sup S$ {{qed}}
:420px {{begin-eqn}}  {{eqn | l = \map \Area {ABC}       | r = \frac 1 2 h c       | c = Area of Triangle in Terms of Side and Altitude }} {{eqn | r = \frac 1 2 h \paren {p + q} }} {{eqn | r = \frac 1 2 a b \paren {\frac p a \frac h b + \frac h a \frac q b} }} {{eqn | r = \frac 1 2 a b \paren {\sin \alpha \cos \beta + \cos \alpha \sin \beta}       | c = {{Defof|Sine of Angle}} and {{Defof|Cosine of Angle}} }} {{eqn | r = \frac 1 2 a b \, \map \sin {\alpha + \beta}       | c = Sine of Sum }} {{eqn | r = \frac 1 2 a b \sin C }} {{end-eqn}} {{qed}}
:400px By definition of sine: :$h = b \sin C$ From Area of Triangle in Terms of Side and Altitude: :$\map \Area {ABC} = \dfrac {a h} 2$ Substituting: :$\map \Area {ABC} = \dfrac {a b \sin C} 2$ {{qed}}
:300px Let $AC$ and $BD$ be intersecting chords of circle $ABCD$. Let the point of intersection be $E$. If $E$ is the center of $ABCD$ the solution is trivial, as $AE = EC = BE = ED$ and so $AE \cdot EC = BE \cdot ED$. Otherwise, let $F$ be the center of $ABCD$. Let $FG$ be drawn perpendicular to $AC$, and $FH$ be drawn perpendicular to $BD$. From Conditions for Diameter to be Perpendicular Bisector, $G$ bisects $AC$ and $H$ bisects $BD$. So $AG = GC$ and $BH = HD$. From Difference of Two Squares we have that $AE \cdot EC + EG^2 = GC^2$. Let us add $GF^2$ to these. So $AE \cdot EC + EG^2 + GF^2 = GC^2 + GF^2$. But from Pythagoras's Theorem we have that: : $GC^2 + GF^2 = CF^2$ : $EG^2 + GF^2 = EF^2$ So: : $AE \cdot EC + EF^2 = CF^2$ Using the same construction, we have that: : $DE \cdot EB + EF^2 = BF^2$ But $BF = CF$ as both are the radius of the circle $ABCD$. That gives us: : $AE \cdot EC + EF^2 = DE \cdot EB + EF^2$ It follows that $AE \cdot EC = DE \cdot EB$ {{Qed}}
Join $A$ with $B$ and $C$ with $D$, as shown in this diagram: :310px Then we have: {{begin-eqn}} {{eqn | l = \angle AEB       | o = \cong       | r = \angle DEC       | c = Two Straight Lines make Equal Opposite Angles }} {{eqn | l = \angle BAE       | o = \cong       | r = \angle CDE       | c = Angles in Same Segment of Circle are Equal }} {{end-eqn}} By Triangles with Two Equal Angles are Similar we have $\triangle AEB \sim \triangle DEC$. Thus: {{begin-eqn}} {{eqn | l = \frac {AE} {EB}       | r = \frac {DE} {EC}       | c =  }} {{eqn | ll= \leadsto       | l = AE \cdot EC       | r = DE \cdot EB       | c =  }} {{end-eqn}} {{qed}}
Let $B = \sup S$. Let $T = \set {x + \xi: x \in S}$. Since $\forall x \in S: x \le B$ it follows that: :$\forall x \in S: x + \xi \le B + \xi$ Hence $\xi + B$ is an upper bound for $T$. If $C$ is the supremum for $T$ then $C \le \xi + B$. On the other hand: :$\forall y \in T: y \le C$ Therefore: :$\forall y \in T: y - \xi \le C - \xi$ Since $S = \set {y - \xi: y \in T}$ it follows that $C - \xi$ is an upper bound for $S$ and so $B \le C - \xi$. So we have shown that $C \le \xi + B$ and $C \ge \xi + B$, hence the result. {{qed}}
=== Proof for Bounded Above === First we show that $\displaystyle \map {\sup_{x \mathop \in S} } {\map f x + c} = c + \map {\sup_{x \mathop \in S} } {\map f x}$: Let $T = \set {\map f x: x \in S}$. Then: {{begin-eqn}} {{eqn | l = \map {\sup_{x \mathop \in S} } {\map f x + c}       | r = \map {\sup_{y \mathop \in T} } {y + c}       | c =  }} {{eqn | r = c + \map {\sup_{y \mathop \in T} } y       | c = Supremum Plus Constant }} {{eqn | r = c + \map {\sup_{x \mathop \in S} } {\map f x}       | c =  }} {{end-eqn}} Next we show that $\displaystyle \map {\sup_{x \mathop \in S} } {\map f x + \map g x} \le \map {\sup_{x \mathop \in S} } {\map f x} + \map {\sup_{x \mathop \in S} } {\map g x}$: Let: :$\displaystyle H = \map {\sup_{x \mathop \in S} } {\map f x}$ :$\displaystyle K = \map {\sup_{x \mathop \in S} } {\map g x}$ Then: :$\forall x \in S: \map f x + \map g x \le H + K$ Hence $H + K$ is an upper bound for $\set {\map f x + \map g x: x \in S}$. The result follows. {{qed}} === Proof for Bounded Below === This follows exactly the same lines. First we show that: :$\displaystyle \map {\inf_{x \mathop \in S} } {\map f x + c} = c + \map {\inf_{x \mathop \in S} } {\map f x}$ Let $T = \set {\map f x: x \in S}$. Then: {{begin-eqn}} {{eqn | l = \map {\inf_{x \mathop \in S} } {\map f x + c}       | r = \map {\inf_{y \mathop \in T} } {y + c}       | c =  }} {{eqn | r = c + \map {\sup_{y \mathop \in T} } y       | c = Infimum Plus Constant }} {{eqn | r = c + \map {\inf_{x \mathop \in S} } {\map f x}       | c =  }} {{end-eqn}} Next we show that: :$\displaystyle \map {\inf_{x \mathop \in S} } {\map f x + \map g x} \ge \map {\inf_{x \mathop \in S} } {\map f x} + \map {\inf_{x \mathop \in S} } {\map g x}$ Let: :$\displaystyle H = \map {\inf_{x \mathop \in S} } {\map f x}$ :$\displaystyle K = \map {\inf_{x \mathop \in S} } {\map g x}$ Then: :$\forall x \in S: \map f x + g \map x \ge H + K$ Hence $H + K$ is a lower bound for $\set {\map f x + \map g x: x \in S}$. The result follows. {{qed}}
From Negative of Infimum is Supremum of Negatives: :$\displaystyle -\inf_{x \mathop \in T} x = \map {\sup_{x \mathop \in T} } {-x} \implies \inf_{x \mathop \in T} x = -\map {\sup_{x \mathop \in T} } {-x}$ Let $S = \set {x \in \R: -x \in T}$. From Negative of Infimum is Supremum of Negatives, $S$ is bounded above. From Multiple of Supremum: :$\displaystyle \map {\sup_{x \mathop \in S} } {z x} = z \, \map {\sup_{x \mathop \in S} } x$ Hence: :$\displaystyle \map {\inf_{x \mathop \in T} } {z x} = -\map {\sup_{x \mathop \in T} } {-z x} = -z \, \map {\sup_{x \mathop \in T} } {-x} = z \, \map {\inf_{x \mathop \in T} } x$ {{qed}}
Construct the altitude from $B$. :200px It can be seen from the definition of sine that: : $\sin A = \dfrac h c$ and $\sin C = \dfrac h a$ Thus: : $h = c \sin A$ and $h = a \sin C$ This gives: : $c \sin A = a \sin C$ So: : $\dfrac a {\sin A} = \dfrac c {\sin C}$ Similarly, constructing the altitude from $A$ gives: :$\dfrac b {\sin B} = \dfrac c {\sin C}$ {{qed}}