Source — Euclid's Elements, encoded as an rrxiv paper

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III.3 Proposition III.3

If in a circle a straight line through the centre bisect a straight line not through the centre, it also cuts it at right angles; and if it cut it at right angles, it also bisects it.

Proof

Let

A B

be a chord not through the centre

E

, and

C D

a line through

E

meeting

A B

at

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

at

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

.

lines 74–74 in main.tex