PROP. XI.—Problem.
To find a third proportional to two given lines (X, Y ).
Sol.—Draw any two lines AC, AE making an angle. Cut off AB equal X, BC equal Y , and AD equal Y . Join BD, and draw CE parallel to BD, then DE is the third proportional required.
Dem.—In the triangle CAE, BD is parallel to CE; therefore AB : BC :: AD : DE [ii.]; but AB is equal to X, and BC, AD each equal to Y . Therefore X : Y :: Y : DE. Hence DE is a third proportional to X and Y .
Another solution can be inferred from Proposition viii. For if AD, DC in that Proposition be respectively equal to X and Y , then DB will be the third proportional. Or again, if in the diagram, Proposition viii., AD = X, and AC = Y , AB will be the third proportional. Hence may be inferred a method of continuing the proportion to any number of terms.
Exercises.
1. If AOΩ be a triangle, having the side AΩ greater than AO; then if we cut off AB = AO, draw BB′ parallel to AO, cut off BC = BB′, &c., the series of lines AB, BC, CD, &c., are in continual proportion.
2. AB − BC : AB :: AB : AΩ. This is evident by drawing through B′ a parallel to AΩ.
PROP. XII.—Problem.
To find a fourth proportional to three given lines (X, Y, Z).