45. In a given circle it is required to inscribe a tri- angle so that two sides may pass through two given points, and the third side he parallel to a given straight line.
Let A and B be the given points, and CD the given straight line. Suppose PMN to be the required triangle inscribed in the given circle.
Draw NE parallel to AB join EM, and produce it if necessary to meet AB at F.
If the point F were known the problem might be considered solved. For ENM is a known angle, and therefore the chord EM is known in magnitude. And then, since F is a known point, and EM is a known magnitude, the position of M becomes known.
We have then only to shew how F is to be determined. The angle MEN is equal to the angle MFA (I. 29). The angle MEN is equal to the angle MPN (III. 21). Hence MAF and BAP are similar triangles (VI. 4). Therefore MA is to AF as BA is to AP. Therefore the rectangle MA, AP is equal to the rectangle BA, AF(VI. 16). But since J is a given point the rectangle MA, AP is known; and AB is known: thus AF is determined.
