Because the straight line AD which meets the two
straight lines BC, EF, makes the alternate angles BAD,
ADC equal to one another, [Construction.
EF is parallel to BC. [I. 27.
Wherefore the straight line EAF is drawn through the given point A, parallel to the given straight line BC. q.e.f.
PROPOSITION 32. THEOREM.
If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are together equal to two right angles.
Let ABC be a triangle, and let one of its sides BC be produced to D : the exterior angle ACD shall be equal to the two interior and opposite angles CAB,ABC; and the three interior angles of the triangle, namely, ABC, BCA, CAB shall be equal to two right angles.
Through the point C draw
CE parallel to AB. [1.31.
Then, because AB is par-
allel to CE, and AC falls on
them, the alternate angles
BAC, ACE are equal. [I. 29.
Again, because AB is parallel to CE, and BD falls on
them, the exterior angle ECD is equal to the interior and
opposite angle ABC. [I. 29.
But the angle ACE was shewn to be equal to the angle
BAC;
therefore the whole exterior angle ACD is equal to the
two interior and opposite angles CAB, ABC. [Axiom 2.
To each of these equals add the angle ACB;
therefore the angles ACD, ACB are equal to the three
angles CBA, BAC, ACB. [Axiom 2.
But the angles ACD, ACB are together equal to two right
angles ; [I. 13.
Therefore also the angles CBA, BAC, ACB are together
equal to two right angles. [Axiom 1.
Wherefore, if a side of any triangle &c. q.e.d.
