Let there be a segment ABG enclosed by a straight line and an orthotome, and from A and parallel to the diameter let ZA be led, and from G and tangent to the orthotome at G let GZ be led. If in fact some line parallel to AZ be led in triangle ZAG, the line drawn will be cut by the orthotome and AG will be cut by the line drawn in the same ratio, but the segment of AG at A will be in the same part of the ratio as the segment of the led line at A.
Let some line DE parallel to AZ be drawn, and first let DE bisect AG. And so since ABG is an orthotome and BD is drawn parallel to the diameter, but AD and DG are equal, then the tangent to the orthotome at B will be parallel to AG. Again, since DE is parallel to the diameter, and GE was drawn from G and tangent to the orthotome at G, but DG is parallel to the tangent at B, therefore EB is equal to BD. Thus, AD has the same ratio to DG which DB has to BE. And so it has been proved if the drawn line bisects AG.
Otherwise some other line KL will be parallel to AZ. And so we must prove that AK has the same ratio to KG which KQ has to QL. For since BE is equal to BD, IL is also equal to KI. Therefore LK has the same ratio to KI which AG has to DA. But KI also has to KQ the same ratio which DA has to AK. For this was proved previously. Thus KQ has the same ratio to QL which AK has to KG. And so the proposed has been proved.