Euclid's Elements Book I

1. Let AB be the given straight line, C the given triangle and D the given rectilineal angle. It is required to apply to the straight line AB, in an angle equal to the angle D, a parallelogram equal to the triangle C. (Every point can be dragged.)
2. Let the parallelogram BEFG be constructed equal to the triangle C, in the angle EBG equal to the angle D, and let it be placed so that BE is in a straight line with AB.

   Proposition I.42

   Proposition I.42: To construct, in a given rectilineal angle, a parallelogram equal to a given triangle.
3. Let FG be drawn through to H, and through A let AH be drawn parallel to either BG or EF.

   Postulate 2Proposition I.31

   Postulate 2: To produce a finite straight line continuously in a straight line.

   Proposition I.31: Through a given point to draw a straight line parallel to a given straight line.
4. Let HB be joined. Since the angles AHF, HFE are equal to two right angles, the angles BHG, GFE are less than two right angles; so HB, FE when produced will meet. Let them meet at K.

   Proposition I.29Postulate 5

   Proposition I.29: A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the interior angles on the same side equal to two right angles.

   Postulate 5: That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.
5. Through K let KL be drawn parallel to either EA or FH, and let HA, GB be produced to the points L, M.

   Proposition I.31

   Proposition I.31: Through a given point to draw a straight line parallel to a given straight line.
6. Then HLKF is a parallelogram, HK is its diameter, and LB, BF are the complements; therefore LB is equal to BF.

   Proposition I.43

   Proposition I.43: In any parallelogram the complements of the parallelograms about the diameter are equal to one another.
7. But BF is equal to the triangle C; therefore LB is also equal to C. And the angle GBE is equal to the angle ABM, and so to the angle D. Therefore the parallelogram LB, equal to the triangle C, has been applied to the straight line AB in the angle ABM equal to the angle D.

   Common Notion 1Proposition I.15

   Common Notion 1: Things which are equal to the same thing are also equal to one another.

   Proposition I.15: If two straight lines cut one another, they make the vertical angles equal to one another.

Being what it was required to do.