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

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III.20 Proposition III.20

In a circle the angle at the centre is double of the angle at the circumference, when the angles have the same circumference as base.

Proof

Let

∠ B E C

be the angle at the centre

E

and

∠ B A C

the angle at the circumference, both subtending the arc

B C

. Join

A E

and produce to

F

on the far side of the circle. In

△ E A B

:

E A = E B

(radii), so by I.5,

∠ E A B = ∠ E B A

. By I.32 (exterior angle of a triangle),

∠ B E F = ∠ E A B + ∠ E B A = 2 \cdot ∠ E A B

. Similarly in

△ E A C

:

∠ C E F = 2 \cdot ∠ E A C

. Adding (Common Notion 2):

∠ B E C = ∠ B E F + ∠ C E F = 2 \cdot (∠ E A B + ∠ E A C) = 2 \cdot ∠ B A C

. (Heath handles the case where

A

lies on the arc opposite

B C

from

E

via a subtraction instead of an addition; the result is the same.)

Figure

$Proposition III.20. The central angle $\angle AOB$ is twice the inscribed angle $\angle APB$ subtending the same arc $AB$. Corollary: all inscribed angles on the same arc are equal.$

Proposition III.20. The central angle

∠ A O B

is twice the inscribed angle

∠ A P B

subtending the same arc

A B

. Corollary: all inscribed angles on the same arc are equal.

lines 74–74 in main.tex

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