1st. The rhomboid (a), whose adjacent sides are not equal and whose angles are not right angles. 2d. The rhombus, or lozenge (b), whose sides are all equal. 3d. The rectangle (c), whose angles are all equal and therefore right angles. 4th. The square (d), whose sides are all equal and whose angles are all equal. The square is at once a rhombus and a rectangle. PROPOSITION XXX.-THEOREM. 104. In every parallelogram, the opposite angles are equal, and the opposite sides are equal. Let ABCD be a parallelogram. 1st. The opposite angles B and D, contained by parallel lines lying in opposite directions, are equal (61); and for the same reason the opposite angles A and C are equal. 2d. Draw the diagonal AC. Since AD and BC are parallel, the alternate angles CAD and ACB are equal (49), and since DC and AB are parallel, the alternate angles ACD and CAB are equal. Therefore, the triangles ADC and CBA are equal (78), and the sides opposite to the equal angles are equal, namely, AD BC, and DC=AB. = 105. Corollary I. A diagonal of a parallelogram divides it into two equal triangles. 106. Corollary II. If one angle of a parallelogram is a right angle, all its angles are right angles, and the figure is a rectangle. PROPOSITION XXXI.-THEOREM. 107. If the opposite angles of a quadrilateral are equal, or if its opposite sides are equal, the figure is a parallelogram. 1st. Let the opposite angles of the quadrilateral ABCD be equal, or AC and B = D. Then, by adding equals, we have A+B=C+D; A B D C But the sum of the therefore, each of the sums A + B and C+D is equal to one-half the sum of the four angles. four angles is equal to four right angles (100); therefore, A + B is equal to two right angles, and the lines AD and BC are parallel (56). In like manner it may be proved that AB and CD are parallel. Therefore the figure is a parallelogram. A D 2d. Let the opposite sides of the quadrilateral ABCD be equal, or BC = AD and AB = DC. Then, drawing the diagonal AC, the triangles ABC, ACD are equal (80); therefore, the angles CAD and ACB are equal, and the lines AD and BC are parallel (54). Also since the angles CAB and A CD are equal, the lines AB and DC are parallel. Therefore ABCD is a parallelogram. B PROPOSITION XXXII.-THEOREM. 108. If two opposite sides of a quadrilateral are equal and parallel, the figure is a parallelogram. Let the opposite sides BC and AD of the quadrilateral ABCD be equal and parallel. Draw the diagonal AC. The alternate angles CAD and ACB are equal (49), and hence the triangles ADC and CBA are equal (76). Therefore, the sides AB and CD are equal and the figure is a parallel.gram (107). B D PROPOSITION XXXIII.-THEOREM. 109. The diagonals of a parallelogram bisect each other. Let the diagonals AC, BD of the parallelogram ABCD intersect in E; then, AE = =EC and ED = EB. B E For, the side AD and the angles EAD, ADE, of the triangle EAD, are respectively equal to the side CB and the angles ECB, EBC of the triangle ECB; hence these triangles are equal (78), and the sides respectively opposite the equal angles are equal, namely, AE EC and ED = EB. = D 110. Corollary I. The diagonals of a rhombus ABCD bisect each other at right angles in E. For, since AD CD and AE EC, ED is perpendicular to AC (41). 111. Corollary II. The diagonals of a rhombus bisect its opposite angles. For, in each of the isosceles triangles ADC, ABC, BCD, DAB, the line drawn from the vertex to the middle of the base bisects the vertical angle (87). E PROPOSITION XXXIV.-THEOREM. 112. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. Let the diagonals of the quadrilateral ABCD bisect each other in E. Then, the triangles AED and CEB are equal (76), and the angles EAD, ECB, respect A D ively opposite the equal sides, are equal. There fore AD and BC are parallel (54). In like manner AB and DC are shown to be parallel, and the figure is a parallelogram. 113. Corollary. If the diagonals of a quadrilateral bisect each other at right angles, the figure is a rhombus. 4. PROPOSITION XXXV.-THEOREM. 114. The diagonals of a rectangle are equal. Let ABCD be a rectangle; then its diagonals, AC and BD, are equal. For, the right triangles ABC and DCB are equal (76); therefore, AC BD. = 115. Corollary I. The diagonals of a square are equal, and, since the square is also a rhombus, they bisect each. other at right angles (110), and also bisect the angles of the square (111). 116. Corollary II. A parallelogram is a rectangle if its diagonals are equal. 117. Corollary III. A quadrilateral is a square, if its diagonals are equal and bisect each other at right angles. 118. Scholium. The rectangle, being a species of parallelogram, has all the properties of a parallelogram. The square, being at once a parallelogram, a rectangle and a rhombus, has the properties of all these figures. PROPOSITION XXXVI.-THEOREM. 119. Two parallelograms are equal when two adjacent sides and the included angle of the one are equal to two adjacent sides and the included angle of the other. Let AC, A'C'', have AB = A'B', ADA'D', and the angle BAD = B'A'D'; then, these parallelograms are equal. D C D' C' Α B A' B' For they may evidently be applied the one to the other so as to coincide throughout. 120. Corollary. Two rectangles are equal when they have equal bases and equal altitudes. APPLICATIONS. PROPOSITION XXXVII.-THEOREM. 121. If a straight line drawn parallel to the base of a triangle bisects one of the sides, it also bisects the other side; and the portion of it intercepted between the two sides is equal to one-half the base. B F D E C Let DE be parallel to the base BC of the triangle ABC, and bisect the side AB in D; then, it bisects the side AC in E, and DE = BC. 1st. Through D suppose DF to be drawn parallel to AC. In the triangles ADE, DBF, we have AD DB, and the angles adjacent to these sides equal, namely DAE = BDF, and ADE = DBF (51); therefore these triangles are equal (78), and AE DF. Also, since DECF is a parallelogram, DF EC (104); and hence AE= EC. 2d. The triangles ADE and BDF being equal, we have DE = BF, and in the parallelogram DECF we have DE = FC; therefore BF FC. Hence Fis the middle point of BC, and DE = BC. = = = 122. Corollary I. The straight line DE, joining the middle points of the sides AB, AC, of the triangle ABC, is parallel to third side BC and is equal to one-half of BC.. For, the straight line drawn through D parallel to BC, passes through E (121), and is therefore identical with DE. Consequently, also, DE={BC. 123. Corollary II. The straight line drawn parallel to the bases of a trapezoid, bisecting one of the non-parallel sides, also bisects the opposite side. B E A D F Let ABCD be a trapezoid, BC and AD its parallel bases, E the middle point of AB, and let EF be drawn parallel to BC or AD; then, Fis the middle of DC. For, draw the diagonal AC, intersecting EF in H. Then in the triangle ABC, EH is drawn through the middle of AB parallel to BC; therefore His the middle of AC. In the triangle ACD, HF is drawn through the middle of AC parallel to AD; therefore F is the middle of DC. 124. Corollary III. In a trapezoid, the straight line joining the middle points of the non-parallel sides is parallel to the bases, and is equal to one-half their sum. |