on B PROP. 36. THEOR. Parallelograms on equal bases and between the same parallels are equal to one another. Given ABCD, EFGH, H parallelograms equal bases BC, FG, and between the same parallels BG, AH. F G To prove | AC = | EG. Join EB and HC. :: BC=FG (given and FG=EH (1.34), :. BC = EH; and BC is also || EH, :. EB is and || CH (1. 33). :: EBCH is a parallelogram. ДАС | BH, because they are on same base BC, and between same parallels BC and AH (1. 35). Similarly, I BH = | EG (I. 35). :: AC = | EG (Ax. 1). Q.E.D. 106. Find a rectangle equal in area to a given parallelogram. 107. Assuming that the area of a rectangle is the product of its length by its breadth, show that the area of any parallelogram is the product of its base by its altitude. 108. Parallelograms which have equal bases and equal altitudes are equal to one another. 109. To bisect a parallelogram by a straight line drawn through a given point in one of its sides, also by a straight line through any given point. 110. To make a rhombus equal to a given parallelogram. 111. Two parallelograms are between the same parallels, and the base of one is a multiple of the base of the other. Show that the area of the first is the same multiple of the area of the other. PROP. 37. THEOR. Triangles on the same base and between the same parallels are equal to one another. Given ABC, DBC, two E triangles on the base BC, and between the parallels BC, AD. To prove A ABC A DBC. B с Produce AD both ways; through B draw BE|| AC (I. 31), and through C draw CF || BD. Then EBCA, DBCF are parallelograms on the same base BC, and between the same parallels BC and EF. :: | EC = ( BF (I. 35). A ABC is half ] EC (1. 34), and A DBC is half | BF, .:. Δ ABC ADBC (Ax. 7). Q.E.D. PROP. 38. THEOR. H the CE (1. 31), and through F draw FH || ED. GC and EH are parallelograms, and they are equal to one another (1. 36); A ABC is half GC (1. 34), and A DEF is half | EH, .:. Δ ABC = A DEF (Ax. 7). Q.E.D. ។ same A B PROP. 39. THEOR. base BC. с Join EC. :: ABC and AEBC are on the same base BC, and between the same parallels BC and AE; :: AABC = AEBC (I. 37). But A ABC = ADBC (given), . :: EBC = ADBC, which is impossible. :: AE is not || BC, and no other line through A but AD can be parallel to BC. :. AD is || BC. Q.E.D. 112. If a straight line is drawn from the vertex of a triangle to bisect the base, the triangle is divided into two parts equal in area. 113. In I. 39, show that AE must meet BD or BD produced. 114. If from two points in a straight line equal lines are drawn on the same side of it, making equal angles with the given line, the line which joins their other extremities is parallel to the given line. 115. If of the four triangles into which the diagonals divide a quadrilateral, two, whose bases are opposite sides of the quadrilateral, are equal, then the other two sides of the quadrilateral are parallel. 116. Two equal triangles are on the same base and on opposite sides of it; prove that the line joining their vertices is bisected by the base or base produced. 117. Equal triangles between the same parallels are on equal ases. D PROP. 40. THEOR. Equal triangles on equal bases in the same straight line, and on the same side of the line, are between the same parallels. Given Δ ABC ADEF on the equal bases BC, same straight line BF. To prove AD | BF. B E F If AD is not || BF, through A draw AG || BF, meeting ED at G (I. 31). Join GF. :: triangles ABC, GEF are on equal bases BC, EF, and between the same parallels BF and AD, .:. Δ ABC AGEF (I. 38). But A ABC = ADEF (given), :: : AGEF = A DEF, which is impossible. :: AG is not || BF, and no other line through A but AD can be parallel to BF. :. AD is || BF. Q.E.D. 118. Triangles between the same parallels and on equal bases, but so that the base of the one is on the same line as the vertex of the other, are equal. 119. The four triangles into which a parallelogram is divided by its diagonals are equal in area. 120. The two triangles formed by taking any part of the diagonal of a parallelogram, or of the diagonal produced as common base, and the two opposite angular points of the parallelogram as vertices, are equal to one another. 121. To construct a triangle equal to a quadrilateral, with one side equal to a side of the quadrilateral. 122. To bisect a triangle by a straight line drawn through a given point in one of its sides. PROP. 41. THEOR. If a parallelogram and a triangle be on the same base and between the same parallels, the parallelogram shall be double the triangle. Given | ABCD and EBC E on the same base BC, and between the same parallels BC and AE. To prove [ BD double of AEBC. Join AC. :: | BD is double of A ABC (1. 34), and A-ABC A EBC (1. 37), :: | BD is double of A EBC. Q.E.D. Note. This proposition is important: it shows that when the area of a parallelogram or of a rectangle is found, the area of a triangle on same base and of same altitude, being equal to half the area of the parallelogram, can also be found. The area of a parallelogram is the product of its base by its altitude (Ex. 107); .. the area of a triangle is half the product of its base by its altitude. 123. If a point be taken within a parallelogram and joined with its angular points, the sum of the two triangles whose bases are two opposite sides of the parallelogram is equal to half of the parallelogram. 124. The same is true if the point be taken anywhere between the two sides, taken as bases, produced ; if not between these two lines, then sum of the two triangles is changed to difference. 125. Construct a triangle equal to a given quadrilateral, whose base shall be in the same straight line with a side of the quadrilateral, and whose vertex shall be one of the angular points of the quadrilateral. 126. If four straight lines be drawn through the angular points of a parallelogram parallel to the diagonals, they shall form a parallelogram double the area of the original parallelogram. |