PROPOSITION 5. PROBLEM.
To describe a circle about a given triangle.
Let ABC be the given triangle: it is required to describe a circle about ABC.
Bisect AB, AC at the points D, E; [I. 10.
from these points draw DF, EF, at right angles to AB,AC; [I. 11.
DF, EF, produced, will meet one another;
for if they do not meet they are parallel,
therefore AB, AC, which are at right angles to them are parallel; which is absurd:
let them meet at F, and join FA; also if the point F be not in BC, join BF, CF.
Then, because AD is equal to BD [Construction.
and DF is common, and at right angles to AB, therefore the base FA is equal to the base FB. [I. 4.
In the same manner it may be shewn that FC is equal to FA.
Therefore FB is equal to FC; [Axiom 1.
and FA, FB, FC are equal to one another.
Therefore the circle described from the centre F, at the distance of any one of them, will pass through the extremities of the other two, and will be described about the triangle ABC.
Wherefore a circle has been described about the given triangle. q.e.f.
Corollary. And it is manifest, that when the centre of the circle falls within the triangle, each of its angles is less than a right angle, each of them being in a segment greater than a semicircle; and when the centre is in one of the sides of the triangle, the angle opposite to this side, being in a semicircle, is a right angle; and when the centre
