In a given triangle to inscribe a circle
be the given triangle; thus it is required to inscribe a circle in the triangle ABC
Let the angles ABC
be bisected by the straight lines BD
], and let these meet one another at the point D
; from D
be drawn perpendicular to the straight
Now, since the angle ABD
is equal to the angle CBD
, and the right angle BED
is also equal to the right angle BFD
are two triangles having two angles equal to two angles and one side equal to one side, namely that subtending one of the equal angles, which is BD
common to the triangles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26]
therefore DE is equal to DF.
For the same reason DG is also equal to DF.
Therefore the three straight lines DE
to one another; therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will pass also through the remaining points, and will touch the straight lines AB, BC, CA, because the angles at the points E, F, G
For, if it cuts them, the straight line drawn at right angles to the diameter of the circle from its extremity will be found to fall within the circle : which was proved absurd; [III. 16
] therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will not cut the straight lines AB, BC, CA;
therefore it will touch them, and will be the circle inscribed in the triangle ABC. [IV. Def. 5]
Let it be inscribed, as FGE
Therefore in the given triangle ABC
the circle EFG
has been inscribed. Q. E. F.