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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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