PROBLEM IV. 135. Through a given point on a given line, to draw a perpendicular to this line. GIVEN, the line AB, and the point C in it. EXERCISES. 5. Solve 134 without an equilateral triangle. 6. Divide a given angle into four equal parts. 3. Find a point in a line at a given distance from a given point. When is the problem impossible? 4. Find a point in a line equally distant from two given. points without the line. 5. From a given point without a given line, draw a line making with the given line an angle equal to one-half a right angle. 6. A is a point without a given line, BC. Construct an isosceles triangle, having A as vertex, whose base shall lie along BC, and be equal to a given sect. 11. If a right-angle triangle have one acute angle double the other, the hypothenuse is double one side. 132. GIVEN, the sect AB. REQUIRED, to bisect it. CONSTRUCTION. On AB describe an equilateral triangle ABC, by By 134, bisect the angle ACB by the line meeting AB in D. Then AB shall be bisected in D. PROOF. In the triangles ACD and BCD (124. Triangles are congruent if two sides and the included angle are equal in each.) AB is bisected at D. 137. COROLLARY I. The line drawn to bisect the angle at the vertex of an isosceles triangle, also bisects the base, and is perpendicular to it. 138. COROLLARY II. The line drawn from the vertex of an isosceles triangle to bisect the base, is perpendicular to it, and also bisects the vertical angle. EXERCISES. 12. Construct a right-angled isosceles triangle on a given sect as hypothenuse. PROBLEM VI. 139. From a point without a given line, to drop a perpendicular upon the line. The. GIVEN, the line AB, and the point C without it. REQUIRED, to drop from C a perpendicular upon AB. CONSTRUCTION. Take any point, D, on the other side of AB from C, and, by 102, from the center C, with radius CD, describe the arc FDG, meeting AB at F and G. By 136, bisect FG at H. Join CH. CH shall be 1 AB (using for the words "perpendicular to "). PROOF. Because in the triangles CHF and CHG, by construction, CG, and HF = HG, and CH is common, CF = :. = (129. Triangles with three sides respectively equal are congruent.) FHC, being half the st. FHG, is a rt. ; and CH is perpendicular to AB. 140. COROLLARY. The line drawn from the vertex of an isosceles triangle perpendicular to the base, bisects it, and also bisects the vertical angle. EXERCISES. 13. Instead of bisecting FG, would it do to bisect the angle FCG? CHAPTER V. INEQUALITIES. 141. The symbol > is called the Sign of Inequality, and a> b means that a is greater than b; so a <b means that a is less than b. THEOREM X. 142. An exterior angle of a triangle is greater than either of the two opposite interior angles. HYPOTHESIS. Let the side AB of the triangle ABC be produced PROOF. By 136, bisect BC in H. By 100 and 101, join AH, and produce it to F; making, by 133, HF = AH. By 100, join BF. Then, in the triangles AHC and FHB, by construction, (124. Triangles are congruent if two sides and the included angle are equal in each.) * HCA = 4 HBF. Now, CBD > & HBF, (85. The whole is greater than its part.) :. CBD > HCA. Similarly, if CB be produced to G, it may be shown that ..the vertical ¥ CBD > 4 BAC. THEOREM XI. 143. Any two angles of a triangle are together less than a straight angle. B C HYPOTHESIS. Let ABC be any ▲. PROOF. Produce CA to D. Then X BAD > B, 1 D (142. An exterior angle of a triangle is greater than either opposite interior angle.) 144. COROLLARY I. No triangle can have more than one right angle or obtuse angle. 145. COROLLARY II. There can be only one perpendicular from a point to a line. |