Proposition III.2 — If on the circumference of a circle two points be taken at r…

Proof

Let

A

B

be on the circle with centre

E

(III.1). Suppose for contradiction that some point

F

on the chord

A B

lies outside the circle; then

E F > E A

. Join

E A

E B

. By I.5, the base angles of the isoceles

△ E A B

are equal:

∠ E A B = ∠ E B A

. By I.16, the exterior angle at any interior point

F

A B

is greater than either remote interior angle; pursuing the inequalities (Heath's argument) forces

E F < E A

for

F

inside

A B

, contradicting the assumption. Hence every point of

A B

lies within the circle.

Knowledge graph · drag to pan, scroll to zoom, click a node to navigate

Full neighborhood

Depends on (4)

III.1Proposition III.1To find the centre of a given circle.
I.5Proposition I.5In isosceles triangles the angles at the base are equal to one another; and if the equal straight lines be produced…
I.16Proposition I.16In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and…
I.15Definition I.15A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among…

Discussion

No replications, contradictions, or comments registered yet for this claim.

Replicate or annotate this claim

Replicate to register a fresh attempt; contradict, extend, or comment otherwise. Authors can post a claim-retraction with the reason taxonomy from RRP-0020.