◀ ▲ ▶Branches / Geometry / Elements-euclid / Book--3-circles / Proof: By Euclid

Proof: By Euclid

(related to Proposition: 3.33: Construction of Segment on Given Line Admitting Given Angle)

• So the [angle] $C$ is surely either acute, a right angle, or obtuse.
• First of all, let it be acute.
  • And, as in the first diagram (from the left), Let (angle) $BAD$, equal to angle $C$, have been constructed on the straight line $AB$, at the point. A (on it) [Prop. 1.23].
  • Thus, $BAD$ is also acute.
  • Let $AE$ have been drawn, at right angles to $DA$ [Prop. 1.11].
  • And let $AB$ have been cut in half at $F$ [Prop. 1.10].
  • And let $FG$ have been drawn from point $F$, at right angles to $AB$ [Prop. 1.11].
  • And let $GB$ have been joined.
  • And since $AF$ is equal to $FB$, and $FG$ (is) common, the two (straight lines) $AF$, $FG$ are equal to the two (straight lines) $BF$, $FG$ (respectively).
  • And angle $AFG$ (is) equal to [angle] $BFG$.
  • Thus, the base $AG$ is equal to the base $BG$ [Prop. 1.4].
  • Thus, the circle drawn with center $G$, and radius $GA$, will also go through $B$ (as well as $A$).
  • Let it have been drawn, and let it be (denoted) $ABE$.
  • And let $EB$ have been joined.
  • Therefore, since $AD$ is at the extremity of diameter $AE$, (namely, point) $A$, at right angles to $AE$, the (straight line) $AD$ thus touches the circle $ABE$ [Prop. 3.16 corr.] .
  • Therefore, since some straight line $AD$ touches the circle $ABE$, and some (other) straight line $AB$ has been drawn across from the point of contact $A$ into circle $ABE$, angle $DAB$ is thus equal to the angle $AEB$ in the alternate segment of the circle [Prop. 3.32].
  • But, $DAB$ is equal to $C$.
  • Thus, angle $C$ is also equal to $AEB$.
  • Thus, a segment $AEB$ of a circle, accepting the angle $AEB$ (which is) equal to the given (angle) $C$, has been drawn on the given straight line $AB$.
• And so let $C$ be a right angle.
  • And let it again be necessary to draw a segment of a circle on $AB$, accepting an angle equal to the right-[angle] $C$.
  • Let the (angle) $BAD$ [again] have been constructed, equal to the right angle $C$ [Prop. 1.23], as in the second diagram (from the left).
  • And let $AB$ have been cut in half at $F$ [Prop. 1.10].
  • And let the circle $AEB$ have been drawn with center $F$, and radius either $FA$ or $FB$.
  • Thus, the straight line $AD$ touches the circle $ABE$, on account of the angle at $A$ being a right angle [Prop. 3.16 corr.] .
  • And angle $BAD$ is equal to the angle in segment $AEB$.
  • For (the latter angle), being in a semicircle, is also a right angle [Prop. 3.31].
  • But, $BAD$ is also equal to $C$.
  • Thus, the (angle) in (segment) $AEB$ is also equal to $C$.
  • Thus, a segment $AEB$ of a circle, accepting an angle equal to $C$, has again been drawn on $AB$.
• And so let (angle) $C$ be obtuse.
  • And Let (angle) $BAD$, equal to ($C$), have been constructed on the straight line $AB$, at the point $A$ (on it) [Prop. 1.23], as in the third diagram (from the left).
  • And let $AE$ have been drawn, at right angles to $AD$ [Prop. 1.11].
  • And let $AB$ have again been cut in half at $F$ [Prop. 1.10].
  • And let $FG$ have been drawn, at right angles to $AB$ [Prop. 1.10].
  • And let $GB$ have been joined.
  • And again, since $AF$ is equal to $FB$, and $FG$ (is) common, the two (straight lines) $AF$, $FG$ are equal to the two (straight lines) $BF$, $FG$ (respectively).
  • And angle $AFG$ (is) equal to angle $BFG$.
  • Thus, the base $AG$ is equal to the base $BG$ [Prop. 1.4].
  • Thus, a circle of center $G$, and radius $GA$, being drawn, will also go through $B$ (as well as $A$).
  • Let it go like $AEB$ (in the third diagram from the left).
  • And since $AD$ is at right angles to the diameter $AE$, at its extremity, $AD$ thus touches circle $AEB$ [Prop. 3.16 corr.] .
  • And $AB$ has been drawn across (the circle) from the point of contact $A$.
  • Thus, angle $BAD$ is equal to the angle constructed in the alternate segment $AHB$ of the circle [Prop. 3.32].
  • But, angle $BAD$ is equal to $C$.
  • Thus, the angle in segment $AHB$ is also equal to $C$.
  • Thus, a segment $AHB$ of a circle, accepting an angle equal to $C$, has been drawn on the given straight line $AB$.
• (Which is) the very thing it was required to do.¹

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References

Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"

Footnotes