Marks' first lessons in geometry / In two parts. Objectively presented, and designed for the use of primary classes in grammar schools, academies, etc.

by Bernhard Marks

Because the angle f is equal to the angle c, the line f d will fall upon the line c a, and the point d will be somewhere in the line c a.

Then, because the point d is in the two lines, b a and c a, it must be in their intersection, or upon the point a.

And, as the two triangles coincide throughout their whole extent, they are equal in all their parts.

That is, the angle a is found to be equal to the angle d; the side b a to the side e d; the side c a to the side f d; and the area of the triangle a b c to the area of the triangle d e f.

PROPOSITION XV. THEOREM.

DEMONSTRATION.

We wish to prove that

The opposite sides of any parallelogram are equal.

Let the figure a b c d be a parallelogram; then will the sides a b and c d be equal to each other; likewise the sides a d and b c.