For if we designate the sides by a, b, and c, as usual, and let CD = d and PB = x, then
(CP)2 = a2 - x2
= b2 - (c - x)2.
∴ a2 - x2 = b2 - c2 + 2cx - x2.
∴ x = (a2 - b2 + c2) / 2c.
∴ (CP)2 = a2 - ((a2 - b2 + c2) / 2c)2.
But CP · d = ab.
∴ d = 2abc / √(4a2c2 - (a2 - b2 + c2)2).
This is not available at this time, however, because the Pythagorean Theorem has not been proved.
These two propositions are merely special cases of the following general theorem, which may be given as an interesting exercise:
If ABC is an inscribed triangle, and through C there are drawn two straight lines CD, meeting AB in D, and CP, meeting the circle in P, with angles ACD and PCB equal, then AC × BC will equal CD × CP.
Fig. 1 is the general case where D falls between A and B. If CP is a diameter, it reduces to the second figure given on [page 249]. If CP bisects ∠ACB, we have Fig. 3, from which may be proved the proposition given at the foot of [page 248]. If D lies on BA produced, we have Fig. 2. If D lies on AB produced, we have Fig. 4.
This general proposition is proved by showing that ⧌ADC and PBC are similar, exactly as in the second proposition given on [page 249].
These theorems are usually followed by problems of construction, of which only one has great interest, namely, To divide a given line in extreme and mean ratio.