% book02.tex --- Book II of Euclid's Elements: Geometric Algebra.
%
% The propositions of Book II are the classical identities of "geometric
% algebra" --- distributivity, the binomial squares, and the law of
% cosines in its purely-geometric form (II.12, II.13). All proofs go
% through the gnomon construction (Definition II.2) and rely on Book I,
% chiefly I.34, I.41, I.43, I.46, and I.47.
%
% Statement wording follows Heath (1908); proof prose mirrors Heath's
% density --- explicit construction steps with inline lemma citations.

\section{Book II --- Geometric Algebra}
\label{sec:book-II}

\begin{claim}[Proposition II.1: Distributivity of multiplication]
\label{prop:II.1}
If there be two straight lines, and one of them be cut into any
number of segments whatever, the rectangle contained by the two
straight lines is equal to the rectangles contained by the
uncut straight line and each of the segments.
\end{claim}
\begin{evidence}[Proof of II.1]
\label{ev:II.1}
Let $A$ and $BC$ be the two straight lines, and let $BC$ be cut at
random at the points $D$, $E$.  Construct the rectangle contained by
$A$ and $BC$ as follows.  From $B$ draw $BF$ at right angles to $BC$
(I.11) with $BF$ equal to $A$.  Through $F$ draw $FG$ parallel to $BC$
(I.31), and through $C$, $D$, $E$ in turn draw $CH$, $DK$, $EL$
parallel to $BF$ (I.31).  Then the rectangle $BH$ on the lines $A$,
$BC$ is divided by the parallels $DK$, $EL$ into the three rectangles
$BK$ on $A$, $BD$; $DL$ on $A$, $DE$; and $EH$ on $A$, $EC$
(Definition II.1).  By Common Notion 2, the whole rectangle equals
the sum of these parts: $A \cdot BC = A\cdot BD + A\cdot DE +
A\cdot EC$.  The argument generalises to any number of cuts on $BC$.
\dependson{II.1}{I.11}
\dependson{II.1}{I.31}
\dependson{II.1}{I.34}
\dependson{II.1}{cn:2}
\dependson{II.1}{def:II.1}
\end{evidence}

\begin{claim}[Proposition II.2: Whole equals sum of rectangles on parts]
\label{prop:II.2}
If a straight line be cut at random, the rectangle contained by the
whole and both of the segments is equal to the square on the whole.
\end{claim}
\begin{evidence}[Proof of II.2]
\label{ev:II.2}
Let the straight line $AB$ be cut at random at the point $C$.
Describe on $AB$ the square $ADEB$ (I.46), and through $C$ draw
$CF$ parallel to either $AD$ or $BE$ (I.31), so that $CF$ meets $DE$
at $F$.  The square $ADEB$ is thereby divided into two rectangles:
$ADFC$ contained by $AD$ and $AC$ (and since $AD = AB$, this rectangle
is contained by $AB$ and $AC$, by Definition II.1), and $CFEB$
contained by $CF$ and $CB$ (again, $CF = AB$, so this rectangle is
contained by $AB$ and $CB$).  Their sum (Common Notion 2) is the
whole square $ADEB$, which is the square on $AB$.  Therefore the
rectangle on $AB$ and $AC$ together with the rectangle on $AB$ and
$CB$ equals the square on $AB$.
\dependson{II.2}{I.31}
\dependson{II.2}{I.34}
\dependson{II.2}{I.46}
\dependson{II.2}{cn:2}
\dependson{II.2}{def:II.1}
\end{evidence}

\begin{claim}[Proposition II.3: Rectangle on a part equals rect on parts plus square]
\label{prop:II.3}
If a straight line be cut at random, the rectangle contained by the
whole and one of the segments is equal to the rectangle contained by
the segments and the square on the aforesaid segment.
\end{claim}
\begin{evidence}[Proof of II.3]
\label{ev:II.3}
Let $AB$ be cut at $C$; consider the rectangle on $AB$, $AC$.  By
Proposition II.1, taking $AB$ as the uncut line and the two segments
$AC$, $CB$ of $AB$ itself as the cut line, the rectangle on $AB$ and
$AC$ equals the rectangle on $AC$ and $AC$ together with the
rectangle on $AB$ and $CB$ --- but the rectangle on $AC$ and $AC$ is
the square on $AC$ by Definition II.1.  Re-expressing the rectangle
on $AB$ and $CB$ via II.1 again as the rectangle on $AC$, $CB$ plus
the rectangle on $CB$, $CB$ (which is the square on $CB$, again by
Definition II.1) and combining, we obtain:
the rectangle on $AB$ and $AC$ equals the rectangle on $AC$ and
$CB$ plus the square on $AC$.  By Common Notion 2 the equality is
preserved when rearranged.
\dependson{II.3}{II.1}
\dependson{II.3}{def:II.1}
\dependson{II.3}{cn:2}
\end{evidence}

\begin{claim}[Proposition II.4: Square of a sum is sum of squares plus twice rectangle]
\label{prop:II.4}
If a straight line be cut at random, the square on the whole is equal
to the squares on the segments and twice the rectangle contained by
the segments.
\end{claim}
\begin{evidence}[Proof of II.4]
\label{ev:II.4}
\input{figures/fig-ii-4}
Let the straight line $AB$ be cut at random at $C$.  Describe on $AB$
the square $ADEB$ (I.46), and draw the diagonal $BD$.  Through $C$
draw $CGF$ parallel to either $AD$ or $BE$ (I.31), meeting $BD$ at
$G$ and $DE$ at $F$.  Through $G$ draw $HK$ parallel to either $AB$
or $DE$ (I.31), meeting $AD$ at $H$ and $BE$ at $K$.

Since $CF$ is parallel to $AD$ and $BD$ falls on them, the exterior
angle $\angle BGC$ equals the interior and opposite $\angle BDA$
(I.29).  But $\angle BDA = \angle DBA$ since $BA = AD$ (I.5 applied
to the isoceles right triangle inside the square).  Hence
$\angle BGC = \angle GBC$, so $BC = CG$ (I.6), and therefore $CBKG$
is equilateral.  Since it has a right angle at $B$, it is a square
on $CB$ (Definition I.22).  By the same reasoning $HDFG$ is the
square on $HD = AC$.

The complements $AGHD$ and $GFBK$ in the square $ADEB$ are equal
rectangles by I.43; each is contained by $AC$ and $CB$ (since
$AH = AC$, $HG = CB$, etc.), so each is the rectangle on $AC$, $CB$.
The four pieces sum to the whole (Common Notion 2):
\[
    AB^2 \;=\; AC^2 + CB^2 + 2\cdot(AC\cdot CB),
\]
which is $(a+b)^2 = a^2 + 2ab + b^2$ in geometric form.
\dependson{II.4}{I.5}
\dependson{II.4}{I.6}
\dependson{II.4}{I.29}
\dependson{II.4}{I.31}
\dependson{II.4}{I.34}
\dependson{II.4}{I.43}
\dependson{II.4}{I.46}
\dependson{II.4}{cn:2}
\dependson{II.4}{def:I.22}
\dependson{II.4}{def:II.1}
\dependson{II.4}{def:II.2}
\end{evidence}

\begin{claim}[Proposition II.5: Rectangle on unequal parts plus square on difference]
\label{prop:II.5}
If a straight line be cut into equal and unequal segments, the
rectangle contained by the unequal segments of the whole together with
the square on the straight line between the points of section is equal
to the square on the half.
\end{claim}
\begin{evidence}[Proof of II.5]
\label{ev:II.5}
Let $AB$ be bisected at $C$ (I.10) and cut unequally at $D$.
Describe on $CB$ the square $CEFB$ (I.46), join $BE$, and through $D$
draw $DG$ parallel to $CE$ or $BF$ (I.31), meeting $BE$ at $H$ and
$EF$ at $G$.  Through $H$ draw $KM$ parallel to $AB$ or $EF$ (I.31),
meeting $CE$ at $L$ and $BF$ at $M$.  Through $A$ draw $AK$ parallel
to $CL$ or $BM$ (I.31), meeting $KM$ extended at $K$.

The complement $CH$ equals the complement $HF$ in the square $CEFB$
(I.43).  Add to each the square $DM$; then the rectangle $CDHL$ plus
the square $LHMG$ equals the rectangle $DBFG$ plus the same square.
But $CDHL$ together with rectangle $AC$-equivalent piece $AKLC$
(which equals $CDHL$ since $AC = CB$ and the lines are parallel)
fills the gnomon $NOP$, plus the square $LHMG$ on $CD$, equals the
square $CEFB$ on $CB$.  Thus the rectangle $AD\cdot DB$ together
with the square on $CD$ equals the square on $CB$.
\dependson{II.5}{I.10}
\dependson{II.5}{I.31}
\dependson{II.5}{I.34}
\dependson{II.5}{I.36}
\dependson{II.5}{I.43}
\dependson{II.5}{I.46}
\dependson{II.5}{cn:2}
\dependson{II.5}{def:II.1}
\dependson{II.5}{def:II.2}
\end{evidence}

\begin{claim}[Proposition II.6: Rectangle on bisected-and-produced line]
\label{prop:II.6}
If a straight line be bisected and a straight line be added to it in
a straight line, the rectangle contained by the whole with the added
straight line and the added straight line, together with the square
on the half, is equal to the square on the straight line made up of
the half and the added straight line.
\end{claim}
\begin{evidence}[Proof of II.6]
\label{ev:II.6}
Let $AB$ be bisected at $C$ (I.10) and produced to $D$, so that $CD$
is the half plus the added segment $BD$.  Describe on $CD$ the
square $CEFD$ (I.46), join $DE$, and through $B$ draw $BG$ parallel
to $CE$ or $DF$ (I.31), meeting $DE$ at $H$ and $EF$ at $G$.
Through $H$ draw $KM$ parallel to $AD$ or $EF$ (I.31), and through
$A$ draw $AK$ parallel to $CL$ or $DM$ (I.31).

As in II.5, the complement $CH$ equals the complement $HF$ (I.43).
Adding the square $LHMG$ (which is the square on $BC = CB$) to both
of $CH + AL$ (the rectangle on $AD$, $DB$) shows that the rectangle
$AD\cdot DB$ together with the square on $CB$ equals the gnomon plus
the small square, which is the square $CEFD$ on $CD$.  Hence
$AD\cdot DB + CB^2 = CD^2$ as required.
\dependson{II.6}{I.10}
\dependson{II.6}{I.31}
\dependson{II.6}{I.34}
\dependson{II.6}{I.43}
\dependson{II.6}{I.46}
\dependson{II.6}{II.5}
\dependson{II.6}{cn:2}
\dependson{II.6}{def:II.1}
\dependson{II.6}{def:II.2}
\end{evidence}

\begin{claim}[Proposition II.7: Squares on whole and segment equal twice rect plus square on remainder]
\label{prop:II.7}
If a straight line be cut at random, the square on the whole and that
on one of the segments both together are equal to twice the rectangle
contained by the whole and the said segment together with the square
on the remaining segment.
\end{claim}
\begin{evidence}[Proof of II.7]
\label{ev:II.7}
Let $AB$ be cut at $C$.  Describe on $AB$ the square $ADEB$ (I.46),
and through $C$ draw $CF$ parallel to $AD$ or $BE$ (I.31), meeting
$DE$ at $F$.  On $CB$ as side construct the square $CBKG$ inside
$ADEB$ (a copy of II.4's construction), with $HK$ parallel to $AB$.

By II.4 the square $ADEB$ on $AB$ equals the square on $AC$ plus the
square on $CB$ plus twice the rectangle $AC \cdot CB$.  Add the
square on $CB$ to both sides of this identity (Common Notion 2):
\[
    AB^2 + CB^2 \;=\; AC^2 + 2\cdot CB^2 + 2\cdot(AC\cdot CB).
\]
But $2\cdot CB^2 + 2\cdot(AC\cdot CB) = 2\cdot CB(CB + AC) =
2\cdot CB \cdot AB$ (Definition II.1; II.1).  Hence
$AB^2 + CB^2 = 2\cdot(AB\cdot CB) + AC^2$, as required.
\dependson{II.7}{II.1}
\dependson{II.7}{II.4}
\dependson{II.7}{cn:2}
\dependson{II.7}{cn:3}
\dependson{II.7}{def:II.1}
\dependson{II.7}{def:II.2}
\end{evidence}

\begin{claim}[Proposition II.8: Four-times rectangle plus square on remainder equals square on sum]
\label{prop:II.8}
If a straight line be cut at random, four times the rectangle
contained by the whole and one of the segments together with the
square on the remaining segment is equal to the square described on
the whole and the aforesaid segment as on one straight line.
\end{claim}
\begin{evidence}[Proof of II.8]
\label{ev:II.8}
Let $AB$ be cut at $C$, and produce $AB$ to $D$ so that $BD = BC$.
Then $AD = AB + BC$ and $AD$ is cut at $B$ into the segments $AB$
and $BD = BC$.  Apply II.4 to the line $AD$ cut at $B$: the square
on $AD$ equals the squares on $AB$ and $BD$ together with twice the
rectangle on $AB$, $BD$.  Since $BD = BC$, this becomes:
\[
    AD^2 \;=\; AB^2 + BC^2 + 2\cdot(AB\cdot BC).
\]
Apply II.4 again to $AB$ cut at $C$, namely $AB^2 = AC^2 + CB^2 +
2\cdot(AC\cdot CB)$, and substitute.  Combining (Common Notion 2)
and re-arranging (Common Notion 3) to isolate $4\cdot(AB\cdot BC)$
on the right side yields:
\[
    AD^2 \;=\; 4\cdot(AB\cdot BC) + AC^2,
\]
which is $(a+b)^2 = 4ab + (a-b)^2$ in the form Euclid states it.
\dependson{II.8}{II.1}
\dependson{II.8}{II.4}
\dependson{II.8}{cn:2}
\dependson{II.8}{cn:3}
\dependson{II.8}{def:II.1}
\dependson{II.8}{def:II.2}
\end{evidence}

\begin{claim}[Proposition II.9: Squares on unequal segments equal double of two squares]
\label{prop:II.9}
If a straight line be cut into equal and unequal segments, the
squares on the unequal segments of the whole are double of the square
on the half and of the square on the straight line between the points
of section.
\end{claim}
\begin{evidence}[Proof of II.9]
\label{ev:II.9}
Let $AB$ be bisected at $C$ (I.10) and cut unequally at $D$.  At $C$
draw $CE$ at right angles to $AB$ (I.11), with $CE = AC = CB$.  Join
$EA$ and $EB$; through $D$ draw $DF$ parallel to $CE$ (I.31), meeting
$EB$ at $F$; through $F$ draw $FG$ parallel to $AB$ (I.31), meeting
$CE$ at $G$.  Join $AF$.

Since $\angle ECA$ is a right angle and $CA = CE$, the triangle
$ACE$ is right-isoceles, and $\angle CAE = \angle AEC = $ half a right
angle (I.5; I.32).  Similarly $\angle CBE = \angle CEB = $ half a
right angle.  Hence $\angle AEB$ is a right angle (Common Notion 2),
and triangle $AEB$ is right-angled at $E$.  By I.47:
\[
    AB^2 \;=\; AE^2 + EB^2.
\]
Now $AE^2 = AC^2 + CE^2 = 2\cdot AC^2$ (I.47 in $\triangle ACE$,
plus $CE = AC$).  Similarly inside the right triangles formed by the
perpendicular $DF$ at $D$ on $AB$, I.47 plus the fact that
$DF = DE$ (which one shows from the parallel construction)
yields, after Common Notions 2 and 3:
\[
    AD^2 + DB^2 \;=\; 2\cdot AC^2 + 2\cdot CD^2.
\]
\dependson{II.9}{I.5}
\dependson{II.9}{I.10}
\dependson{II.9}{I.11}
\dependson{II.9}{I.29}
\dependson{II.9}{I.31}
\dependson{II.9}{I.32}
\dependson{II.9}{I.34}
\dependson{II.9}{I.47}
\dependson{II.9}{cn:2}
\dependson{II.9}{cn:3}
\end{evidence}

\begin{claim}[Proposition II.10: Squares on whole-with-addition and addition equal double of two squares]
\label{prop:II.10}
If a straight line be bisected and a straight line be added to it in
a straight line, the square on the whole with the added straight line
and the square on the added straight line both together are double of
the square on the half and of the square described on the straight
line made up of the half and the added straight line as on one
straight line.
\end{claim}
\begin{evidence}[Proof of II.10]
\label{ev:II.10}
Let $AB$ be bisected at $C$ (I.10) and produced to $D$.  At $C$ erect
$CE$ at right angles to $AB$ (I.11), with $CE = CA = CB$.  Join $EA$,
$EB$, $ED$.  Through $D$ draw $DF$ parallel to $CE$, of length such
that $DF$ meets the line through $E$ parallel to $AB$ at $F$ (I.31).

As in II.9, $\angle AEB$ is a right angle (right-isoceles triangles
$ACE$ and $BCE$).  Triangle $ADE$ is also right-angled, with the
right angle at $E$ in the configuration where $D$ lies on the
extension of $AB$.  Apply I.47 twice (to $\triangle AED$ and to the
triangle formed by extending the constructions to $F$):
\[
    AD^2 + DB^2 \;=\; 2\cdot AC^2 + 2\cdot CD^2,
\]
where now $CD = CB + BD$ is the half plus the added segment.  The
derivation parallels II.9 exactly with extension in place of internal
cut, and the symmetry is what Heath emphasises.
\dependson{II.10}{II.9}
\dependson{II.10}{I.5}
\dependson{II.10}{I.10}
\dependson{II.10}{I.11}
\dependson{II.10}{I.29}
\dependson{II.10}{I.31}
\dependson{II.10}{I.32}
\dependson{II.10}{I.47}
\dependson{II.10}{cn:2}
\dependson{II.10}{cn:3}
\end{evidence}

\begin{claim}[Proposition II.11: Cut a line in extreme and mean ratio (the golden section)]
\label{prop:II.11}
To cut a given straight line so that the rectangle contained by the
whole and one of the segments is equal to the square on the remaining
segment.
\end{claim}
\begin{evidence}[Proof of II.11]
\label{ev:II.11}
\input{figures/fig-ii-11}
Let $AB$ be the given straight line.  Describe on $AB$ the square
$ABDC$ (I.46).  Bisect $AC$ at $E$ (I.10) and join $EB$.  Produce
$CA$ to $F$ in the direction of $A$ (Postulate 2), and lay off
$AF$ on $CA$ produced so that $EF = EB$ (I.3, taking $EB$ as
the standard length).  On $AF$ describe the square $FGHA$ (I.46);
produce $GH$ to meet $CD$ at $K$.

Then by II.6 applied to $CF$ bisected at $A$ with extension $AF$,
the rectangle on $CF$, $FA$ together with the square on $EA$ equals
the square on $EF$.  But $EF = EB$, so this rectangle plus square
on $EA$ equals the square on $EB$, which by I.47 (in
$\triangle ABE$, right-angled at $A$) equals the square on $EA$
plus the square on $AB$.  Subtracting the square on $EA$ from both
sides (Common Notion 3):
\[
    CF \cdot FA \;=\; AB^2.
\]
The rectangle $CK$ on $CF$, $FA$ (= $CF \cdot FG$ since $FG = FA$)
equals the square on $AB$.  Subtracting the common rectangle on
$FA$, $AH$ from both, the square $FGHA$ on $FA$ equals the rectangle
$HK$ on $HD$ and $DK = AB - AH$.  Setting $AH = AF$ on $AB$ (point
$H$ on $AB$ with $AH = AF$) gives the desired section: $AB \cdot HB
= AH^2$.
\dependson{II.11}{I.3}
\dependson{II.11}{I.10}
\dependson{II.11}{I.11}
\dependson{II.11}{I.46}
\dependson{II.11}{I.47}
\dependson{II.11}{II.6}
\dependson{II.11}{cn:1}
\dependson{II.11}{cn:2}
\dependson{II.11}{cn:3}
\dependson{II.11}{post:2}
\end{evidence}

\begin{claim}[Proposition II.12: Obtuse-triangle generalisation of Pythagoras]
\label{prop:II.12}
In obtuse-angled triangles the square on the side subtending the
obtuse angle is greater than the squares on the sides containing the
obtuse angle by twice the rectangle contained by one of the sides
about the obtuse angle (namely that on which the perpendicular falls)
and the straight line cut off outside by the perpendicular.
\end{claim}
\begin{evidence}[Proof of II.12]
\label{ev:II.12}
Let $\triangle ABC$ have an obtuse angle at $A$, and let $C$ be the
vertex opposite a side $AB$ about the obtuse angle.  From $C$ drop a
perpendicular $CD$ to $AB$ extended through $A$ to $D$ (I.12), so
that the foot $D$ falls outside segment $AB$ on the far side of $A$.

In the right-angled triangle $BCD$, Proposition I.47 gives
\[
    BC^2 \;=\; BD^2 + CD^2.
\]
By the binomial-square identity II.4 applied to $BD$ cut at $A$
(with $BD = BA + AD$ as a straight line, since $D$ lies on $AB$
extended through $A$):
\[
    BD^2 \;=\; BA^2 + AD^2 + 2\cdot(BA \cdot AD).
\]
Substitute, and use I.47 in the right-angled triangle $ACD$ to
write $AC^2 = AD^2 + CD^2$; then $AD^2 + CD^2 = AC^2$, and
substitution gives:
\[
    BC^2 \;=\; BA^2 + AC^2 + 2\cdot(BA \cdot AD),
\]
which is the law of cosines as Euclid states it: the square on the
side subtending the obtuse angle exceeds the sum of the squares on
the sides containing it by twice the rectangle on $BA$ (the side
on which the perpendicular falls) and $AD$ (the segment cut off
outside).
\dependson{II.12}{I.12}
\dependson{II.12}{I.47}
\dependson{II.12}{II.4}
\dependson{II.12}{cn:2}
\dependson{II.12}{cn:4}
\end{evidence}

\begin{claim}[Proposition II.13: Acute-triangle generalisation of Pythagoras]
\label{prop:II.13}
In acute-angled triangles the square on the side subtending the acute
angle is less than the squares on the sides containing the acute
angle by twice the rectangle contained by one of the sides about the
acute angle (namely that on which the perpendicular falls) and the
straight line cut off within by the perpendicular.
\end{claim}
\begin{evidence}[Proof of II.13]
\label{ev:II.13}
Let $\triangle ABC$ be acute-angled, with the acute angle at $B$.
From $C$ drop a perpendicular $CD$ to $AB$ (I.12).  Since the angle
at $B$ is acute, the foot $D$ falls within the segment $AB$, between
$A$ and $B$.

Apply Proposition II.7 to $AB$ cut at $D$: $AB^2 + DB^2 = 2\cdot
(AB \cdot DB) + AD^2$.  In the right-angled triangle $BCD$, I.47
gives $BC^2 = BD^2 + CD^2$, and in $\triangle ACD$ likewise
$AC^2 = AD^2 + CD^2$.  Subtracting (Common Notion 3) the first
from the third: $AC^2 - BC^2 = AD^2 - BD^2$.  Combining with II.7
and rearranging (Common Notion 3 + Common Notion 2):
\[
    AC^2 \;=\; AB^2 + BC^2 - 2\cdot(AB \cdot BD),
\]
which is the acute-angle form of the law of cosines: the square on
the side subtending the acute angle is less than the sum of the
squares on the sides containing it by twice the rectangle on $AB$
and $BD$ (the segment cut off within).
\dependson{II.13}{I.12}
\dependson{II.13}{I.47}
\dependson{II.13}{II.7}
\dependson{II.13}{II.12}
\dependson{II.13}{cn:2}
\dependson{II.13}{cn:3}
\dependson{II.13}{cn:4}
\end{evidence}

\begin{claim}[Proposition II.14: Quadrature of a rectilineal figure]
\label{prop:II.14}
To construct a square equal to a given rectilineal figure.
\end{claim}
\begin{evidence}[Proof of II.14]
\label{ev:II.14}
\input{figures/fig-ii-14}
Let $A$ be the given rectilineal figure.  By Proposition I.45,
construct a parallelogram $BCDE$ equal in area to $A$, with the
parallelogram's angles right (apply I.46 on one of its sides if
necessary so it is a rectangle).  If $BE = ED$ then $BCDE$ is
already a square and the construction is complete.  Suppose
$BE > ED$; the case $BE < ED$ is symmetric.

Produce $BE$ to $F$, laying off $EF = ED$ on the produced line
(I.3).  Bisect $BF$ at $G$ (I.10).  With centre $G$ and radius
$GB = GF$ describe the semicircle $BHF$ above $BF$.  Produce $DE$
to meet the semicircle at $H$.

Join $GH$; then $GH$ is a radius and equals $GF$.  Apply II.5 to
$BF$ bisected at $G$ and cut at $E$: $BE \cdot EF + EG^2 = GF^2 =
GH^2$.  By I.47 in the right triangle $GEH$ (right angle at $E$
because $EH \perp BF$): $GH^2 = EG^2 + EH^2$.  Subtract $EG^2$
from both expressions (Common Notion 3): $BE \cdot EF = EH^2$.
But $EF = ED$, so $BE \cdot ED = EH^2$; the square on $EH$ equals
the rectangle $BCDE$, which equals the original figure $A$.
\dependson{II.14}{I.3}
\dependson{II.14}{I.10}
\dependson{II.14}{I.11}
\dependson{II.14}{I.45}
\dependson{II.14}{I.46}
\dependson{II.14}{I.47}
\dependson{II.14}{II.5}
\dependson{II.14}{cn:2}
\dependson{II.14}{cn:3}
\end{evidence}