Book 1 – Proposition 42

To construct, in a given rectilineal angle, a parallelogram equal to a given triangle.
[Desmos graphs can be found here]

Start with a given triangle ABC and a given rectilineal angle D.

The goal is to construct a parallelogram that is equal to the triangle ABC in an angle equal to D.

Let E bisect BC. Join AE. Use Proposition 23 (Construction of an angle on a straight line that is equal to a given angle) to construct an angle CEF that is equal to angle D.

Draw AG through A that is parallel to EC. Draw CG through C that is parallel to EF. These lines can be drawn using Proposition 31 (Construction of a straight line through a given point that is parallel to a given line).

Since FG is parallel to EC and EF is parallel to CG, FECG is a parallelogram.

Now we look at triangles ABE and AEC.
Since BE is equal to EC (as E bisects BC), the two triangles have equal bases.
The two triangles are in the same parallels (BC and AG).
By Proposition 38 (two triangles on equal bases and in the same parallels are equal to one another) triangle ABE is equal to triangle AEC.

Since the triangle ABC is composed of the two equal triangles ABE and AEC, the triangle ABC is double the triangle AEC.

But, by Proposition 41 (if a parallelogram has the same base as a triangle and they are in the same parallels then the parallelogram is double the triangle) the parallelogram FECG is double the triangle AEC since they share the same base (EC) and they are in the same parallels.

Since both the parallelogram FECG and the triangle ABC are double the triangle AEC, the parallelogram FECG is equal to the triangle ABC.
The angle FEC is equal to the angle D.
We have constructed the parallelogram that meets the necessary conditions.