THEOREM XXIV. If a triangle have two angles equal, the sides opposite those angles will be equal, and the triangle will be isosceles. In the triangle ABC, if the angles CAB, CBA, are equal, the opposite sides BC, AC, will be equal. A C D B For, if they are not equal, suppose BC > AC, and take BD = AC. Then in the two triangles BAD and ABC we have the side BD = AC by supposition, the side AB common, and the included angle ABD of the first triangle equal to BAC, the included angle of the second triangle. Hence (T. XX.), the triangle BAD is equal to ABC, that is, a part is equal to the whole, which is impossible (A. VIII.). There is, therefore, no inequality of the sides CB and CA, that is, they are equal, and the triangle is isosceles. THEOREM XXV. When two triangles have the three sides of the one respectively equal to the three sides of the other, the triangles will be identical, and equal in all respects. Let the two triangles ABC, ABD, have their sides respectively equal, namely, AB equal to AB, AC equal to AD, and BC equal to BD, then will these triangles be identical. A C B For, conceive the two triangles to be joined together by their longest equal sides, and draw the line CD; then in the triangle ACD, since AC is equal to AD, we have the angle ACD equal to the angle ADC (T. XXIII.). In like manner, in the triangle BCD, since BC is equal to BD, we have the angle BCD equal to the angle BDC. Hence the angle ACB, which is the sum of ACD and BCD, is equal to the angle ADB, which is the sum of ADC and BDC. Since, then, in the triangle ACB, we have the two sides AC and BC, and their included angle ACB, equal respectively to the two sides AD and BD, and their included angle ADB, of the triangle ADB, it therefore follows that these triangles are identical (T. XX.). THEOREM XXVI. Two right-angled triangles are equal, when the hypotenuse and a side of the one, are respectively equal to the hypotenuse and a side of the other. Suppose BC= EF and AB = DE, then will the two right-angled triangles, ABC and DEF, be equal. C F A G B D E For, if we apply the triangle ABC upon the triangle DEF, so that AB may coincide with its equal DE, the side AC will take the direction of DF, since the angle at A is equal to the angle at D, each being a right angle. If, now, BC coincides with EF, there will be a complete coincidence, and the two triangles will be equal. If possible, suppose BC to take the position EG, so that the triangle ABC may be represented by DEG ; then, since the hypotenuses of the two triangles are equal, we shall have EG EF, that is, two oblique lines are drawn from E, terminating at unequal distances from D, the foot of the perpendicular ED, which is impossible (T. XII.). It is, therefore, absurd to suppose BC to take any position different from EF. Hence, the triangles coincide throughout, and are in all respects equal. = Cor. Two right-angled triangles will also be equal, when the hypotenuse and an acute angle of the one are respectively equal to the hypotenuse and an acute angle of the other. For, since the sum of the two acute angles of any right-angled triangle is equal to a right angle (T. V., C. II.), it follows that the remaining acute angles of the two triangles will be equal. Hence, the two triangles will have two angles and the interjacent side of the one equal, respectively, to two angles and the interjacent side of the other. Consequently, they will be equal (T. XXII.). THEOREM XXVII. The greater side of every triangle is opposite the greater angle, and, conversely, the greater angle is opposite the greater First. Suppose, in the triangle ABC, the angle CBA > CAB, then will the side CA> CB. A D C B For, draw BD, making the angle DBA = DAB, and DA will equal DB (T. XXIV.). To each adding DC, we have DA + DC = DB + DC, but DB + DC is greater than CB (T. VII.); hence, DA + DC, or its equal CA, is greater than CB. Secondly. Suppose, in the triangle ABC, the side CA> CB, then will the angle CBA > CAB. A E C F D B For, through the middle point of the side AB draw the perpendicular DE. Now, since CA> CB, the point C must be on the side of DE opposite A (T. XIII., S.); consequently, the perpendicular must cut CA at some point, as F. Draw FB, and we shall have FB = FA (T. XIII.), and angle FBA = FAB (T. XXIII.); consequently, FBA+FBC, which equals CBA. is greater than CAB. OF QUADRILATERALS. THEOREM XXVIII. If the opposite angles of a quadrilateral are equal, the figure is a parallelogram. A D B C We have the angles A+B+C+D= 4 right angles (T. V., C. IV.), true for all quadrilaterals. Now, if we suppose A =C, and BD, we shall find 2A +2D =4 right angles, or A+D = 2 right angles, hence AB and DC are parallel (T. XVII.). In a similar manner, we find 2A+ 2B 4 right angles, or A+B=2 right angles, and, as before, AD and BC are parallel. Consequently the figure is a parallelogram (D. XXV.). = THEOREM XXIX. If the opposite sides of a quadrilateral are equal, the figure is a parallelogram. If we suppose AB= DC, and AD = BC, and draw the diagonal BD, we shall have the three sides of the triangle ABD respectively equal to the three sides of the triangle CDB, consequently they are equal, (T. XXV.), and the angle ABD = CDB, which are alternate interior angles in reference to the sides AB and DC, hence AB and DC are parallel (T. XVII., C.). The equality of the two triangles also gives the angle ADB = CBD, which are alternate interior angles in reference to the sides AD and BC, hence these sides are parallel. Consequently the figure is a parallelogram (D. XXV.). THEOREM XXX. If two opposite sides of a quadrilateral are equal and allel, the figure is a parallelogram. A D B par C In the quadrilateral ABCD, suppose AB equal and parallel to DC, then will the figure be a parallelogram. For, drawing BD, the two triangles ABD, CDB, have the common side BD, and AB of the first equal to CD of the second; also, the included angles ABD, CDB, equal, being alternate angles with reference to the parallels AB, DC. Hence these triangles are equal (T. XX.), and the angle ADB is equal to CBD, but these are alternate angles with reference to AD and BC, consequently AD and BC are parallel, and the figure is a parallelogram. THEOREM XXXI. The opposite angles of a parallelogram are equal; so also are the opposite sides equal. D C Since the figure is a parallelogram, we have AB and DC parallel, also AD and BC parallel; hence, if we draw the diagonal BD, we shall have the angle ABD = CDB, being alternate interior angles in reference to the parallels AB and DC (T. XVIII., C.). For a similar reason A B 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 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 ABAD, 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. |