Part. En. Let ABC, DEF be the two triangles, having the sides BA, AC equal to the sides ED, DF respectively, and having also the angle B = the angle E; it is required to prove that the angle C is equal or supplementary to the angle F; and that when the angle C is equal to the angle F, the triangles are equal in all respects. Proof. The contained angle A must be either equal or unequal to the contained angle D. If A=D, as in Fig. I, then, by Theorem 5, the triangles are equal in all respects; and the angle C is equal to the angle F. If A is not equal to D, as in Fig. 2, let the point A be placed on D, and AB along DE, then, because AB = DE, (Hyp.) the point B will coincide with the point E; and because the angle at B = the angle at E, (Hyp.) therefore the line BC will lie along EF, and the point C will fall on EF as G: and because AC = DF, (Hyp.) therefore DG=DF, and therefore the angle DFE = the angle DGF; (Th. 6.) but DGF is supplementary to DGE, that is, to ACB, and therefore the angle Fis supplementary to the angle C. Q. E. D. COR. Hence the triangles are equal in all respects (1) If the two angles given equal are right angles or obtuse angles. For then the remaining angles must be acute, and therefore cannot be supplementary, and must therefore be equal by the Theorem, and therefore the triangles must be equal in all respects. (2) If the angles opposite to the other two equal sides are both acute, or both obtuse, or if one of them is a right angle. (3) If the side opposite the given angle in each triangle is not less than the other given side. For then the given angles must be the greater of the two, and therefore the remaining angles must be both acute, and therefore cannot be supplementary, and must therefore be equal, by the Theorem, and therefore the triangles must be equal in all respects. EXERCISES ON THEOREMS OF EQUALITY. The general method to be adopted in the solution of theorems of equality is the following. Examine fully the statement of the question; see what is included among the data : what lines and angles are given equal by hypothesis. Then see what is required to be proved, what lines or angles have to be proved to be equal. It may follow from the properties proved of a single triangle; or it may depend on the equality of a pair of triangles. In the latter case examine the triangles of which they form corresponding parts, and see whether the data are sufficient to prove these triangles equal. If the data are sufficient, the solution is effected by comparing the triangles, and shewing the required equality of the lines and angles ; if not, the data must be used to establish results, which in their turn can be used to establish the conclusion required. The beginner will do well to arrange his proofs in the manner shewn in the example, giving references in the margin. EXERCISES. Ex. (1). The lines which hisect the angles at the base of an isosceles triangle, and meet the opposite sides, are equal. Let ABC be an isosceles triangle. Data. AB=AC, and the angles at B and C (Hyp.)) Therefore the base CE=the base BD. Ex. (2). The bisectors of the three angles of a triangle will meet in one point. Let ABC be a triangle, and let the bisectors of the angles ABC, ACB be BO, CO, meeting in 0; then the Theorem will be proved if we can shew that AO is the bisector of the angle BAC. Let perpendiculars OP, OQ, OR be drawn to OQC=OPC, (Constr.)) OC common. Similarly from the triangles OPB, ORB, it follows that OP=OR; therefore OR=OQ; and therefore the right-angled triangles OQA, ORA have the hypotenuse and one side of the one equal to the hypotenuse and one side of the other, and are therefore equal in all respects by Theorem 20, Cor. 1. Therefore the angle OAQ=the angle OAR, that is, OA is the bisector of the angle BAC. EXERCISES FOR SOLUTION. 1. OA and OB are any two equal lines, and AB is joined; shew that AB makes equal angles with OA and OB. 2. If the bisectors of the equal angles B, C of an isosceles triangle meet in 0, shew that OBC is also an isosceles triangle. 3. The line drawn to bisect the vertical angle of an isosceles triangle will also bisect the base, and be perpendicular to it. 4. The lines joining the middle points of the sides of an isosceles triangle to the opposite extremities of the base will be equal to one another. 5. The line drawn from the vertex of an isosceles triangle to bisect the base will cut it at right angles, and bisect the vertical angle. 6. Prove that the lines which bisect the sides of a triangle and are perpendicular to them meet in one point. 7. The perpendiculars let fall from the extremities of the base of an isosceles triangle upon the opposite sides will be equal, and will make equal angles with the base. 8. The perpendicular let fall from the vertex of an isosceles triangle to the base, will bisect the base and the vertical angle. 9. If two exterior angles of a triangle be bisected by straight lines which meet in 0, prove that the perpendiculars. from 0 on the sides or sides produced of the triangle are, equal to one another. EXERCISES ON THEOREMS OF INEQUALITY. The line that joins the vertex to the middle point of the base of a triangle is less than half the sum of the two sides. Let D be the middle point of AC, then is BD less than half the sum of AB, BC. Proof. Produce BD to B', making DB'=DB. Join AB'. Then since the two triangles BDC, B'DA have two sides BD, DC and the included angle BDC of the one respectively equal to the two sides B'D, DA and the included angle B'DA of the other, therefore (Theorem 5) the base BC=the base AB'; B'A+AB>B'B, (Th. 12.) .. AB+BC>B'B, which is twice BD, that is, BD is less than half the sum of BC and BA. BI but EXERCISES FOR SOLUTION. 1. Prove that any one side of a four-sided figure is less than the sum of the other three sides. 2. Prove that the sum of the lines which join the opposite angles of any four-sided figure is together greater than the sum of either pair of opposite sides of the figure. 3. Prove that the sum of the diagonals of a quadrilateral figure is less than the sum of the four lines which can be drawn to the angles from any other point than the intersection of the diagonals. |