Proposition III.3 — If in a circle a straight line through the centre bisect a s…

Proof

Let

A B

be a chord not through the centre

E

, and

C D

a line through

E

meeting

A B

F

. Suppose

C D

bisects

A B

, so

A F = F B

. Join

E A

E B

. In

△ E A F

and

△ E B F

E A = E B

(radii),

A F = F B

(given),

E F

common. By I.8 the triangles are congruent, so

∠ E F A = ∠ E F B

, and by I.13 both are right. Conversely, if

C D ⊥ A B

F

, then in the right triangles

△ E A F

and

△ E B F

we have

E A = E B

and

E F

common, with right angles at

F

; by I.4 (SAS variant) or I.26 (ASA),

A F = F B

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Depends on (6)

III.1Proposition III.1To find the centre of a given circle.
I.4Proposition I.4If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight…
I.8Proposition I.8If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they…
I.13Proposition I.13If a straight line set up on a straight line make angles, it will make either two right angles or angles equal to two…
I.26Proposition I.26If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely either…
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…

Required by (dependents) (7)

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