Let triangle ABC be the given triangle, and angle YDZ the given rectilinear angle.

Bisect line BC at point E, and join line AE. Construct the angle CEF on the straight line CE at the point E on it equal to the angle YDZ. Draw line AG through point A parallel to line CE, and draw line CG through point C parallel to line EF.

Then polygon FECG is a parallelogram.

Since line BE equals line CE, therefore the triangle ABE also equals the triangle ACE, for they are on equal bases line BE and line CE and in the same parallels line BC and line AG. Therefore the triangle ABC is double the triangle ACE.

But the parallelogram FECG is also double the triangle ACE, for it has the same base with it and is in the same parallels with it, therefore the parallelogram FECG equals the triangle ABC.

And it has the angle CEF equal to the given angle YDZ.

Therefore the parallelogram FECG has been constructed equal to the given triangle ABC, in the angle CEF which equals angle YDZ.

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