Again, because the straight line BAE meets the parallels AD, EC, the exterior angle BAD is equal to the interior and opposite angle AEC; [I.29.
but the angle BAD has been shewn equal to the angle ACE;
therefore the angle ACE is equal to the angle AEC; [Axiom 1.
and therefore AC is equal to AE. [I. 6.
And, because AD is parallel to EC, [Constr.
one of the sides of the triangle BCE,
therefore BD is to DC as BA is to AE; [VI. 2.
but AE is equal to AC;
therefore BD is to DC as BA is to AC. [V.7.
Next, let BD be to DC as BA is to AC, and join AD: the angle BAC shall be bisected by the straight line AD,
For, let the same construction be made. Then BD is to DC as BA is to AC; [Hypothesis.
and BD is to DC as BA is to AE, [VI. 2.
because AD is, parallel to EC; [Construction.
therefore BA is to AC as BA is to AE; [V. 11.
therefore AC is equal to AE; [V. 9.
and therefore the angle AEC is equal to the angle ACE. [1.5.
But the angle AEC is equal to the exterior angle BAD; [1. 29.
and the angle ACE is equal to the alternate angle CAD; [1. 29.
therefore the angle BAD is equal to the angle CAD;[Ax.l.
that is, the angle BAC is, bisected by the straight line AD.
Wherefore, if the vertical angle &c. q.e.d.
