Construction of Parallelogram on Given Line equal to Triangle in Given Angle

Theorem
On any given line segment, a parallelogram can be constructed in a given angle the same size as any given triangle.

Proof

 * Euclid-I-44.png

Let $$AB$$ be the given line segment, $$C$$ be the given triangle and let $$D$$ be the given angle.

From Construction of Parallelogram Equal to Triangle in Given Angle, construct the parallelogram $$BEFG$$ equal to $$C$$ such that $$\angle EBG = \angle D$$.

Place it so that $$BE$$ is in a straight line with $$AB$$.

Let $$FG$$ be produced to $$H$$, and draw $$AH$$ parallel to $$BG$$.

Then join $$HB$$.

We have that $$HF$$ falls on the parallel lines $$AH$$ and $$EF$$.

So from Parallel Implies Supplementary Interior Angles, $$AHF$$ and $$HFE$$ are supplementary.

So $$\angle BHG + \angle GFE$$ is less than two right angles.

From the Fifth Postulate, $$HB$$ and $$FE$$ will meet if produced. So let them meet at $$K$$.

Draw $$KL$$ parallel to $$EA$$, and produce $$HA$$ and $$GB$$ to the points $$L$$ and $$M$$.

Then $$HLKF$$ is a parallelogram, and from Complements of Parallelograms are Equal, $$ABML$$ has the same area as $$BEFG$$.

But $$BEFG$$ has the same area as $$\triangle C$$, and so by common notion 1, so does $$ABML$$.

From Two Straight Lines make Equal Opposite Angles, $$\angle GBE = \angle ABM$$, while $$\angle GBE = \angle D$$.

Therefore $$ABML$$ is the required parallelogram.