EXERCISES ON TRIANGLES. The theorems respecting triangles, which follow from the elementary properties of triangles proved above, are of two kinds; which announce respectively the equality, or the inequality of lines and angles. Theorems of equality depend on Theorems 7, 8, 10, 14, 15, 16, 18, 20, and theorems of inequality on Theorems 9, 11, 12, 13, 17, 19, given above. We shall give a few examples of these theorems with demonstrations, and add some exercises which may be left to the student's ingenuity to prove. They require an application of these general theorems to the special data of the theorem proposed. 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. 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 (in Theorems 7, 8, 10) of a single triangle; or it may depend on the equality of a pair of triangles. 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 examples, giving references in the margin. Theorems of Equality. I. The lines which bisect 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 bisected by BD, CE. Proof. In the triangles AEC, ADB and angle ACE-angle ABD: Data & Th. 8. Therefore the base CE= the base BD, E B Th. 15. 2. The bisectors of the three angles of a triangle meet in one point. Let ABC be a triangle and let the bisectors of the angles ABC, ACB be BO, CO, meeting in O; 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 the three sides BC, CA, AB. Proof. In the triangles OQC, OPC we have OCQ=OCP Data. OC common. Therefore OQ= OP by Theorem 15. R Similarly from the triangles OPB, ORB, it follows that OP=OR; therefore OR OQ; = And therefore the right-angled triangles OQA, ORA have the hypothenuse and one side of the one equal to the hypothenuse and one side of the other, and are therefore equal in all respects by Theorem 20, Cor. 2. Therefore the angle OAQ= the angle OAR, that is, OA is the bisector of the angle BAC. W. G. D EXERCISES ON TRIANGLES. I. Theorems of Equality. 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 O, shew that OBC is also an isosceles triangle. 3. If ABC is an isosceles triangle and A is double of either B or C, shew that A is a right angle. 4. If ABC is an isosceles triangle and A is half of either B or C, shew that A is two-fifths of a right angle. 5. Find the angle between the lines that bisect the angles at the base of the triangle in the last question. 6. The perpendiculars let fall from the extremities of the base of an isosceles triangle on the opposite sides will include an angle supplementary to the vertical angle of the triangle. 7. The line drawn to bisect the vertical angle of an isosceles triangle will also bisect the base, and be perpendicular to it. 8. 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. 9. 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. 10. 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. II. The perpendicular let fall from the vertex of an isosceles triangle to the base, will bisect the base and the vertical angle. 12. If two exterior angles of a triangle be bisected by straight lines which meet in O, prove that the perpendiculars from O on the sides or sides produced of the triangle are equal to one another. 13. Prove that the lines which bisect the sides of a triangle and are perpendicular to them meet in one point. Theorems of Inequality. I. If P is any point within the triangle ABC, prove that PA+ PB <CA + CB, but the angle APB>the angle ACB. Proof. Produce AP to meet BC in Q. Then AC+ CQ>AQ (Th. 13), and therefore AC+CB>AQ+ QB ; also PQ+QB> PB (Th. 13), and therefore AQ+QB>AP+PB; much more then are AC+ CB>AP+PB. And the angle APB>AQB, and AQB>ACB (Th. 7, Cor.); much more then is APB>ACB. 2. 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, ADB have two sides BD, DC and the included angle BDC of the one respectively equal to the two sides B'D, DA B' B THEOREM 15. If two triangles are equiangular to one another and have a side of the one equal to the corresponding side of the other, then triangles will be equal in all respects. Let ABC, DEF be the two triangles, in which the angles A, B, C are respectively equal to D, E, F; and one side BC equal to the corresponding side EF. Then shall the triangles be equal in all respects. Proof. For if the triangle ABC be conceived (Ax. 10) as placed on the triangle DEF, so that the side BC coincides with the equal side EF, then BA will fall on ED, since the angle B the angle E; and CA will fall on FD, since the angle C the angle F = = Therefore A will fall on D, and the triangles will coincide, and are therefore equal in all respects: that is, AB=DE, AC=DF, and the area of the triangle ABC=the area of the triangle DEF. THEOREM 16. If two sides and the included angle of one triangle are respectively equal to two sides and the included angle of another, the triangles will be equal in all respects. Let the two sides BA, AC of the triangle BAC be respectively equal to the two sides ED, DF of the triangle |