T H E E L E M E N T S |
Book I Proposition 32 | |
Therefore the remaining angles BAD, ABD are equal to one right angle. [I.32] But the angle ABF is also right; therefore the angle ABF is equal to the angles BAD, ABD. Let the angle ABD be subtracted from each; therefore the angle DBF which remains is equal to the angle BAD in the alternate segment of the circle. |
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