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AB has been drawn across from the point of contact at A; therefore
the angle DAB is equal to the angle constructed in the alternate segment AHB of the circle. [III.32]
But
the angle BAD is equal to the angle at C.
Therefore
the angle in the segment AHB is also equal to the angle at C.
Therefore
on the given straight line AB the segment AHB of a circle has been described admitting an angle equal to the angle at C.
Being what it was required to do.
