therefore the triangle BDE is to the triangle ADE as the triangle CDE is to the triangle ADE; [V. 11.
that is, the triangles BDE and CDE have the same ratio to the triangle ADE,
Therefore the triangle BDE is equal to the triangle CDE. [V. 9.
And these triangles are on the same base DE and on the same side of it;
but equal triangles on the same base, and on the same side of it, are between the same parallels; [I. 39.
therefore DE is parallel to BC.
Wherefore, if a straight line &c. q.e.d.
PROPOSITION 3. THEOREM.
If the vertical angle of a triangle he bisected by a straight line which also cuts the base, the segments 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 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 vertical angle.
Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD, which meets the base at D: BD shall be to DC as BA is to AC.
Through C draw CE parallel to DA, [I. 31.
and let BA produced meet CE at E.
Then, because the straight line AC meets the parallels AD, EC, the angle ACE is equal to the alternate angle CAD; [I. 29.
but the angle CAD is, by hypothesis, equal to the angle BAD;
therefore the angle BAD is equal to the angle ACE. [Ax. 1.
