|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;
the whole angle BDA is equal to the two angles CDA, DAC.
the exterior angle BCD is equal to the angles CDA, DAC; [I.32]
the angle BDA is also equal to the angle BCD.
|Previous Page Return to Propositions Next Page|