the angle ADB = CBD. Therefore, the two triangles ABD and CDB have a common side BD, and the adjacent angles of the one equal respectively to the adjacent angles of the other, consequently they are equal (T. XXII.). Hence the angle A=C, and ABD+CBD = CDB +ADB, or ABC-CDA. That is, the opposite angles of a parallelogram are equal. The equality of the triangles also gives AB=DC, and AD = BC; that is, the opposite sides of a parallelogram are also equal. Cor. I. The diagonal of a parallelogram divides it into two equal parts. Cor. II. Two parallels included between two other parallels are equal. Cor. III. If one angle of a parallelogram is right, the other three will be right also. THEOREM XXXII. The diagonals of a parallelogram mutually bisect each other, that is, they divide each other into halves. A D E B C In the two triangles EAB and ECD, we have the angle EAB of the first equal to ECD of the second, being alternate angles in reference to the parallels AB and DC. For a like reason the angle EBA is equal to EDC. We also have the interjacent sides AB and DC equal, consequently these triangles are equal (T. XXII.), and AE = EC, BE = ED. Cor. If the four sides of the parallelogram are equal, as in the case of a rhombus, we have AB = AD, and the two triangles AEB and AED will have the three sides of the one equal to the three sides of the other respectively, consequently they will be equal (T. XXV.), and the angle AEB = AED, that is, in a rhombus the diagonals bisect each other at right angles. Scholium. The point E, where the diagonals intersect, is called the centre of the parallelogram. THEOREM XXXIII. The straight line joining the middle points of the oblique sides of a trapezoid, will be parallel to the other sides, and equal to half their sum. E A D C G H F ལའ Suppose EF to join the middle points E and F, of the oblique sides AD and BC of the trapezoid ABCD. Through F draw GH parallel to AD, and meeting DC produced. In the two triangles, FBH and FCG, we have the side FB = FC, the angle FBH equal to its alternate angle FCG, and the angle BFH equal its opposite angle CFG, hence these triangles have two angles and the interjacent side of the one, respectively equal to the two angles and the interjacent sides of the other, and are therefore equal (T. XXII.). Consequently, FGFH; that is, FG equals half of HG. But HG is equal to AD, since the figure AHGD is a parallelogram. Therefore, FG is half of AD, but ED is also half of AD, hence FG and ED are equal and parallel, and the figure EFGD is a parallelogram, and EF is parallel to DC, and also to AB, since DC and AB are parallel. Again, since EF=DG=AH; and CG=HB, by reason of the equality of the two triangles FGC and FHB, we have EF as much greater than DC as it is less than AB. Hence, EF is one half of the sum of DC and AB. Cor. If we suppose the side DC of the trapezoid to be reduced to a point, the trapezoid will then change to a triangle. Hence, the straight line joining the middle points of any two sides of a triangle, will be parallel to the third side, and equal to half of it. THEOREM XXXIV. The four lines joining the middle points of the adjacent sides of a quadrilateral form a parallelogram. H D Ꮐ C F If we draw the diagonals of the quadrilateral, we shall have EF and HG, each parallel to AC, and equal to half of it (T. XXXIII., C.); consequently they are equal and parallel. For the same reason EH and FG are each parallel to BD, and equal to half of it. Therefore the figure EFGH is a parallelogram. A E B Cor. The two lines joining the middle points of the opposite sides of a quadrilateral mutually bisect each other, since they are the diagonals of a parallelogram. ADDITIONAL THEOREMS OF TRIANGLES. THEOREM XXXV. In any triangle if a line be drawn from either angle to the middle of the opposite side: I. When this line is equal to the half side, the angle will be a right angle. II. When this line is greater than the half side, the angle will be acute. III. When this line is less than the half side, the angle will be obtuse. In the triangle ABC, suppose the line CD to be drawn from the angle C to the middle of the side AB. C A D B First. When CD=AD=DB, we have, angle A = DCA, angle B = DCB (T. XXIII.), hence A+B=DCA+DCB; that is, the angle ACB is one half the sum of the three angles of the triangle, and is therefore a right angle (T. V., C. I.). Secondly. When CD > AD or DB, we have angle A > DCA, angle B> DCB (T. XXVII.), hence A+B>DCA+DCB; that is, the angle ACB is less than half the sum of the three angles of the triangle, and it is therefore an acute angle. Thirdly. When CD <AD or DB, we have angle A < DCA, angle B<DCB (T. XXVII.); hence, A+B <DCA+DCB; that is, the angle ACB is greater than half the sum of the three angles of the triangle, and it is therefore an obtuse angle. Scholium. This Theorem affords a very simple method of determining the kind of angle in any given triangle. HEOREM XXXVI. The three lines bisecting the three angles of a triangle, inter sect each other in the same point. Every point in the bisecting line AD is equally distant from AB and AC (T. XIV.). For the same reason every point of the bisecting line BE is equally distant from AB and BC. Hence, the point G, where these two lines intersect, is equally A C E D F B distant from AC and BC; consequently, it is in the line CF which bisects the angle ACB (T. XIV., C. I.). That is, this point is common to the three bisecting lines. Cor. If the sides AB and AC are produced, and the exterior angles are bisected by the lines BF and CF, the point F, where they intersect, will be on the line AF, which bisects the angle at A. A C E K B D Scholium. This common point, equidistant from the three sides of a triangle, is always within the triangle. THEOREM XXXVII. The three perpendiculars bisecting the three sides of a triangle intersect each other in the same point. Every point in the perpendicular DG is equidistant from A and B (T. XIII.). For the same reason every point in the perpendicular EG is equidistant from A and C. Hence, the point G, where A E C C G F E F Ꮐ D B A D B these perpendiculars intersect, is equidistant from B and C; consequently, it is in the perpendicular which bisects the side BC (T. XIII., C. I.). That is, this point is common to the three perpendiculars. Scholium. This common point, equidistant from the three angles of a triangle, may be either within or without the triangle; or, in the case of a right-angled triangle, it will evidently be at the middle point of the hypotenuse. THEOREM XXXVIII. The three lines passing through the three angles of a triangle, and bisecting the opposite sides, will intersect each other in the same point. We will first consider the two lines CD, BE, which intersect at G. Draw HK joining the middle points of GB, GC, and it will be parallel to BC and equal to half of it (T. XXXIII., C.). For the same reason DE, which joins the middle points of AB, AC, will be parallel to BC and equal to half of it; consequently, HK and DE are equal and parallel. Now in the two triangles GHK, GED, we have the angle GHK equal to its alternate angle GED, and GKH equal to its alternate angle GDE, also the side HK=DE; hence, these triangles are equal (T. XXII.), and GH = GE, but GH was taken one half of BG, therefore we have EG one third of EB. For the same reason, DG is one third of DC. If we now consider the two lines CD, AF, we can, by the same kind of demonstration already employed, show that the point where they intersect is one third of FA, measured from F on FA; and one third of DC, measured from D on DC, as before. Hence, the three lines intersect at the same point. Scholium. This common point, situated at one third the distance of the middle points of the sides of a triangle from their opposite angles, is always within the triangle. THEOREM XXXIX. The three lines passing through the three angles of a triangle, perpendicular to the opposite sides, will intersect each other in the same point. Let ABC be the given triangle. Through the vertices draw lines parallel to the opposite sides, thus forming a second triangle DEF. E C F A B D Then will the figures ABCE, ABFC, and ADBC be parallelograms, and we have AE and AD each equal to BC; hence, A is the middle point of ED. For a similar reason, B is the middle point of DF, and C the middle point of EF. From this we see that lines drawn from A, B, C, perpendicularly to BC, AC, AB, will be the same as the three perpendiculars bisecting the sides of the triangle DEF, which perpendiculars we already know must intersect each other in the same point (T. XXXVII.). Hence, three lines passing through the three angles of a triangle perpendicularly to the opposite sides, will intersect each other in the same point. Scholium. This common point may be either within or without the triangle (T. XXXVII.). |