PROPOSITION 23. PROBLEM.
At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle.
Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle: it is required to make at the given point A, in the given straight line AB, an angle equal to the given rectilineal angle DCE.
In CD, CE take any points D, E, and join DE.
Make the triangle AFG the sides of which shall be equal to the three straight lines CD, DE, EC; so that AF shall be equal to CD, AG to CE, and FG to DE. [I. 22.
The angle FAG shall be equal to the angle DCE.
Because FA, AG are equal to DC, CE, each to each, and the base FG equal to the base DE; [Construction.
therefore the angle FAG is equal to the angle DCE. [I. 8.
Wherefore at the given point A in the given straight line AB, the angle FAG has been made equal to the given rectilineal angle DCE. q.e.f.
PROPOSITION 24. THEOREM.
If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them, of the other, the base of that which has the greater angle shall be greater than the base of the other.
Let ABC, DEF be two triangles, which have the two sides AB, AC, equal to the two sides DE, DF, each to each, namely, AB to DE, and AC to DF, but the angle BAC greater than the angle EDF: the base BC shall be
