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

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III.31 Proposition III.31

In a circle the angle in the semicircle is right, that in a greater segment less than a right angle, and that in a less segment greater than a right angle.

Proof

Let

A B

be a diameter of the circle with centre

O

, and

C

any point on the circle other than

A

,

B

. Join

O C

,

A C

,

B C

. Since

O A = O C

(radii),

△ O A C

is isoceles, so by I.5,

∠ O A C = ∠ O C A

. Similarly

△ O B C

is isoceles with

∠ O B C = ∠ O C B

. By I.32, the angles of

△ A B C

sum to two right angles: \[ \angle OAC + \angle OBC + \angle ACB \;=\; \text{two right angles.} \] But

∠ A C B = ∠ O C A + ∠ O C B = ∠ O A C + ∠ O B C

by the isoceles equalities. Substituting,

2 \cdot ∠ A C B =

two right angles, so

∠ A C B

is right. For inscribed angles in segments greater than a semicircle, the arc is less than a semicircle, so by III.20 the inscribed angle is half a central angle less than two right angles — hence less than a right angle. The lesser-segment case is symmetric.

Figure

$Proposition III.31. For any point $C$ on the circle (not at $A$ or $B$), the inscribed angle $\angle ACB$ subtending the diameter $AB$ is a right angle. Proof: by I.5 applied to the two isoceles triangles $OAC$ and $OCB$, then I.32.$

Proposition III.31. For any point

C

on the circle (not at

A

or

B

), the inscribed angle

∠ A C B

subtending the diameter

A B

is a right angle. Proof: by I.5 applied to the two isoceles triangles

O A C

and

O C B

lines 74–74 in main.tex

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