PROPOSITION A. THEOREM.
If the exterior angle of a triangle, made by producing one of its sides, be bisected by a straight line which also cuts the base produced, the segments between the dividing straight line and the extremities of the base shall have the same ratio which the other sides of the triangle have to one another; and if the segments of the base produced have the same ratio which the other sides of the triangle have to one another, the straight line drawn from the vertex to the point of section shall bisect the exterior angle of the triangle.
Let ABC be a triangle, and let one of its sides BA be produced to E; and let the exterior angle CAE be bisected by the straight line AD which meets the base produced at D: BD shall be to DC as BA is to AC.
Through C draw CF parallel to AD, [I. 31.
meeting AB at F.
Then, because the straight line AC meets the parallels AD, FC, the angle ACF is equal to the alternate angle CAD; [1.29.
but the angle CAD is, by hypothesis, equal to the angle DAE;
therefore the angle DAE is equal to the angle ACF. [Ax. 1.
Again, because the straight line FAE meets the parallels AD, FC, the exterior angle DAE is equal to the interior and opposite angle AFC; [I. 29.
but the angle DAE has been shewn equal to the angle ACF;
therefore the angle ACF is, equal to the angle AFC; [Ax. 1.
and therefore AC is equal to AF. [I. 6.
And, because AD is paralled to FC [Construction.
one of the sides of the triangle BCF;
therefore BD is to DC as BA is to AF; [VI. 2.
but AF is equal to AC;
therefore BD is to DC as BA is to AC [V. 7.
