Let line AB be the given straight line, angle XDZ the given rectilinear angle, and triangle STU the given triangle.

It is required to apply a parallelogram equal to the given triangle STU to the given straight line AB in an angle equal to angle XDZ.

Construct the parallelogram BEFG equal to the triangle STU in the angle EBG which equals angle XDZ, and let it be placed so that line BE is in a straight line with line AB.

Draw line FG through to point H, and draw line AH through point A parallel to either line BG or line EF. Join line HB.

Since the straight line HF falls upon the parallels line AH and line EF, therefore the sum of the angles angle AHF and angle HFE equals two right angles. Therefore the sum of the angles BGH and angle GFE is less than two right angles. And straight lines produced indefinitely from angles less than two right angles meet, therefore line HB and line FE, when produced, will meet.

Let them be produced and meet at point K. Draw line KL through the point K parallel to either line EA or line FH. Produce line HA and line GB to the points L and M.

Then polygon HLKF is a parallelogram, line HK is its diameter, and polygon ABGH and polygon MBEK are parallelograms, and line LB and line BF are the so-called complements about line HK. Therefore line LB equals line BF.

But polygon BEFG equals the triangle STU, therefore polygon LMBA also equals triangle STU.

Since the angle GBE equals the angle ABM, while the angle GBE equals angle XDZ, therefore the angle ABM also equals the angle angle XDZ.

Therefore the parallelogram LB equal to the given triangle STU has been applied to the given straight line AB, in the angle ABM which equals angle XDZ.

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