T H E E L E M E N T S |
Book IV Proposition 10 | |
Since, then, BD touches the circle ACD, and DC is drawn across from the point
of contact at D, therefore the angle BDC is equal to the angle DAC in the alternate segment of the circle. [III.32] Since, then, the angle BDC is equal to the angle DAC, let the angle CDA be added to each; therefore the whole angle BDA is equal to the two angles CDA, DAC. But the exterior angle BCD is equal to the angles CDA, DAC; [I.32] therefore the angle BDA is also equal to the angle BCD. |
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