PROPOSITION 42. PROBLEM.
To describe a parallelogram that shall be equal to a given triangle and have one of its angles equal to a given rectilineal angle.
Let ABC be the given triangle, and D the given recti- lineal angle : it is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D.
Bisect BC at E:[I.10.
join AE, and at the point
E, in the straight line EC,
make the angle CEF equal
to D; [I.23.
through A draw AFG
parallel to EC, and through
C draw CG parallel to
EF. [I. 31.
Therefore FECG is a parallelogram. [Definition.
And, because BE is equal to EC, [Construction.
the triangle ABE is equal to the triangle AEC, because
they are on equal bases BE, EC, and between the same
parallels BC, AG. [I. 38.
Therefore the triangle ABC is double of the triangle AEC.
But the parallelogram FECG is also double of the triangle
AEC, because they are on the same base EC, and between
the same parallels EC, AG. [I. 41.
Therefore the parallelogram FECG is equal to the triangle
ABC ; [Axiom 6.
and it has one of its angles CEF equal to the given angle
D. [Construction.
Wherefore a parallelogram FECG has been described equal to the given triangle ABC, and having one of its angles CEF equal to the given angle D. q.e.f.
