Page:The Elements of Euclid for the Use of Schools and Colleges - 1872.djvu/358

334 55. Required the locus of the vertices of all the triangles ABC which stand on a given base AB, and have the side AC to the side BC in a constant ratio.

If the sides AC and BC are to be equal, the locus is

the straight lino which bisects AB at right angles. "We will suppose that the ratio is greater than a ratio of equality; so that AC is to be the greater side. Divide AB at D so that AD is to DB in the given ratio (VI. 10); and produce AB to E, so that AE is to EB in the given ratio. Let P be any point in the required locus; join PD and PE. Then PD bisects the angle APB, and PE bisects the angle between BP and AP produced. Therefore the angle DPE is a right angle. Therefore P is on the circumference of a circle described on DE as diameter.

We have thus shewn that any point which satisfies the assigned conditions is on the circumference of the circle described on DE as diameter; we shall now shew that any point on the circumference of this circle satisfies the assigned conditions.

Let Q be any point on the circumference of this circle, QA shall be to QB in the assigned ratio. For, take O the centre of the circle; and join QO. Then, by construction, AE to EB as AD is to DB, and therefore, alternately, AE to AD as EB is to DB; therefore the sum of AE and AD is to their difference as the sum of EB and DB is to their difference (23); that is, twice DO is to twice DO as twice DO is to twice BO; therefore AO is to DO as DO is