Proposition XLV. Problem.
|(199)||To construct a parallelogram equal to a given rectilinear figure (A B C E D), and having an angle equal to a given one (H).|
Resolve the given rectilinear figure into triangles; construct a parallelogram R Q equal to the triangle B D A (XLIV), and having an angle I equal to a given angle H; on a side of it, R V, construct the parallelogram X V equal to the triangle C B D, and having an angle equal to the given one (XLIV), and so on construct parallelograms equal to the several triangles into which the figure is resolved. L Q is a parallelogram equal to the given rectilinear figure, and having an angle I equal to the given angle H.
Because R V and I Q are parallel the angle V R I together with I is equal to two right angles (XXIX); but V R X is equal to I (const.), therefore V R I with V R X is equal to two right angles, and therefore I R and R X form one right line (XIV); in the same manner it can be demonstrated, that R X and X L form one right line, therefore I L is a right line, and because Q V is parallel to I R the angle Q V R together with V R I is equal to two right angles (XXIX); but I R is parallel to V F, and therefore I R V is equal to F V R (XXIX), and therefore Q V R together with F V R is equal to two right angles, and Q V and F V form one right line (XIV); in the same manner it can be demonstrated of V F and F Y, therefore Q Y is a right line and also is parallel to I L; and because L Y and R V are parallel to the same line X F, I Y is parallel to R V (XXX); but I Q and R V are parallel, therefore L Y is parallel to I Q, and therefore L Q is a parallelogram, and it has the angle I equal to the given angle H, and is equal to the given rectilinear figure A B C E D.
(200) Cor.—Hence a parallelogram can be applied to a given right line and in a given angle equal to a given rectilinear figure, by applying to the given line a parallelogram equal to the first triangle.
Book I: Euclid, Book I (ed. Dionysius Lardner, 11th Edition, 1855)
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