PROPOSITION 5.
About a given triangle to circumscribe a circle.
Let
ABC be the given triangle; thus it is required to circumscribe a circle about the given triangle
ABC.
Let the straight lines
AB,
AC be bisected at the points
D,
E [
I. 10], and from the points
D,
E let
DF,
EF be drawn at right angles to
AB,
AC; they will then meet within the triangle
ABC, or on the straight line
BC, or outside
BC.
First let them meet within at
F, and let
FB,
FC,
FA be joined.
Then, since
AD is equal to
DB, and
DF is common and at right angles, therefore the base
AF is equal to the base
FB. [
I. 4]
Similarly we can prove that
CF is also equal to AF; so that
FB is also equal to
FC;
therefore the three straight lines FA, FB, FC are equal to one another.
Therefore the circle described with centre
F and distance one of the straight lines
FA,
FB,
FC will pass also through the remaining points, and the circle will have been circumscribed about the triangle
ABC.
Let it be circumscribed, as
ABC.
Next, let
DF,
EF meet on the straight line
BC at
F, as is the case in the second figure; and let
AF be joined.
Then, similarly, we shall prove that the point
F is the centre of the circle circumscribed about the triangle
ABC.
Again, let
DF,
EF meet outside the triangle
ABC at
F, as is the case in the third figure, and let
AF,
BF,
CF be joined.
Then again, since
AD is equal to
DB, and
DF is common and at right angles, therefore the base
AF is equal to the base
BF. [
I. 4]
Similarly we can prove that
CF is also equal to AF; so that
BF is also equal to
FC; therefore the circle described with centre
F and distance one of the straight lines
FA,
FB,
FC will pass also through the remaining points, and will have been circumscribed about the triangle
ABC.
Therefore about the given triangle a circle has been circumscribed. Q. E. F.
And it is manifest that, when the centre of the circle falls within the triangle, the angle
BAC, being in a segment greater than the semicircle, is less than a right angle; when the centre falls on the straight line
BC, the angle
BAC, being in a semicircle, is right; and when the centre of the circle falls outside the triangle, the angle
BAC, being in a segment less than the semicircle, is greater than a right angle. [
III. 31]
