[Folio 0055 | p. 49]

PROPOSITION 21.

In a circle the angles in the same segment are equal to one another.

Let ABCD be a circle, and let the angles BAD, BED be angles in the same segment BAED; I say that the angles BAD, BED are equal to one another.

For let the centre of the circle ABCD be taken, and let it be F; let BF, FD be joined.

Now, since the angle BFD is at the centre, and the angle BAD at the circumference ,

and they have the same circumference BCD as base,

therefore the angle BFD is double of the angle BAD. [III. 20]

For the same reason

the angle BFD is also double of the angle BED; therefore the angle BAD is equal to the angle BED.

Therefore etc. Q. E. D.

Dependency Graph

Required for

III.22, III.31, III.32, III.33, III.34, IV.2, IV.5, IV.6, IV.10, IV.11, IV.12, VI.13, VI.25, VI.28, VI.29, VI.30, X.13, X.18, X.22, X.23, X.24, X.25, X.26, X.27, X.28, X.29, X.30, X.31, X.32, X.33, X.34, X.35, X.36, X.37, X.38, X.39, X.40, X.41, X.42, X.43, X.44, X.45, X.46, X.47, X.48, X.49, X.50, X.52, X.53, X.54, X.55, X.56, X.57, X.58, X.59, X.60, X.61, X.62, X.63, X.64, X.65, X.66, X.67, X.68, X.69, X.70, X.71, X.72, X.73, X.75, X.76, X.77, X.78, X.79, X.80, X.81, X.82, X.83, X.84, X.85, X.86, X.87, X.88, X.89, X.90, X.91, X.92, X.93, X.94, X.95, X.96, X.97, X.98, X.99, X.100, X.101, X.102, X.103, X.104, X.105, X.106, X.107, X.108, X.109, X.110, X.111, X.112, X.113, X.114, XI.23, XII.1, XII.2, XII.5, XII.6, XII.7, XII.8, XII.10, XII.11, XII.12, XII.13, XII.14, XII.15, XII.17, XII.18, XIII.6, XIII.11, XIII.12, XIII.13, XIII.14, XIII.16, XIII.17, XIII.18

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See Commentary on Proposition III.21.