AC, & angle ВСЕ. BG, CI. ER AB, AC, be produced to D and E, the qual to ACB, and the angle CBD to the angle BI) take any point F, and make AG equal to AF, and join A B C F G E Because AF is equal to AG, and AB to AC, the two sides AF, AC are equal to the two sides AG and AB, each to each; and they contain the angle FAG common to the two triangles ACF, ABG; therefore (4. 1.), the triangle ACF is equivalent to the triangle ABG; and therefore the angle AFC is equal to the angle AGB: and because the whole AF is equal to the whole AG, of which the parts AB, AC are equal; the remainder BF shall be equal to the remainder CG; but CF is equal (4. 1.) to BG; therefore, the two sides BF, CF are equal to the two CG, BG, each to each; and the angle BFC is equal to the angle CGB; wherefore (4.1.) the triangles are equivalent: therefore, the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG; and, since it has been shewn that the whole angle ABG is equal to the whole angle ACF, the parts of which, the angles CBG, BCF are also equal; the remaining angle ABC is therefore equal to the remaining angle ACB, which are the angles at the base of the triangle ABC; and it has also been proved, that the angle FBC is equal to the angle GCB, which are the angles upon the other side of the base. Therefore, the angles at the base of an isosceles triangle are equal to one another; and, if the equal sides be produced, the angles upon the other side of the base shall be equal. Which was to be proved. D/ Cor. Hence, every equilateral triangle is also equiangular. If two angles of a triangle be equal to one another, the sides also which subtend, or are opposite to, the equal angles, shall be equal to one another. Let ABC be a triangle, having the angle ABC equal to the angle ACB; the side AB is also equal to the side AC. For, if AB be not equal to AC, one of them is greater than the other; let AB be the greater, and from it cut off DB equal to AC the less, and join DC; therefore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides BD, BC are equal to the two sides AC, BC, each to each; and the angle DBC is equal to the angle ACB; therefore (4. 1.), the base DC is equal to the base AB, and the triangle CBD is equivalent to the triangle ACB, the less to the greater; which is absurd. Therefore, AB is not unequal to AC, that is, it is equal to it. Wherefore, if two angles of a triangle be equal to one another, the sides also which subtend, or are opposite to, the equal angles, shall be equal to one another. Which was to be proved. 1 COR. Hence, every equiangular triangle is also equilateral. PROP. VII. THEOR. If two triangles have the three sides of the one equal to the three sides of the other, each to each; the angles opposite the equal sides are also equal. Let the two triangles ABC, DEF, have the three sides equal, that is, AB equal to DE, AC to DF, and BC to EF. The angles opposite the equal sides shall also be equal; that is, the angle BAC to the angle EDF, ABC to DEF, and ACB to DFE. Let the triangles be joined by the longest equal sides BC and EF, having the remaining sides GB equal to DE, and GC to DF, and join AG. Because in the triangle ABG the side AB is equal to the side BG by the hypothesis, the angle BAG is equal to the angle BGA (5.1.); for the same reason, in the triangle ACG the angle CAG is equal to the angle CGA; and therefore, the two angles BAG and CAG, or BAC, are equal to the two angles BGA and CGA, or BGC; then the triangles ABC, GBC have the two AB, AC equal to the two sides GB, GC, and the contained angle BAC equal to the contained angle BGC. And, therefore, the triangles are equivalent (4. 1.); and the angle ABC equal to the angle GBC, or DEF, and the angle ACB to the angle GCB, or DFE. Therefore, if, &c. Q. E. D. PROP. VIII. PROBLEM. To bisect a given finite straight line. Let AB be the given finite straight line; it is required to bisect it. From A, with a radius (Def. 50.) AF greater than half AB, describe the arc DFE; from B, with the same radius, describe the arc DGE, intersecting DFE in D, E, and join DE, intersecting AB at C; AB is bisected at C. Join AD, BD, BE, AE. Because the two triangles ADE, BDE, have the two sides AD, AE, equal to the two sides BD, BE, and the third side BDC; therefore (4. 1.), the triangles are equivalent, and the E third side AC is equal to the third side BC, and the straight line AB is divided into two equal parts in the point C. Which was to be done. PROP. IX. PROB. To draw a straight line at right angles to a given straight line, from a given point in the same. Let AB be the given straight line, and C the given point in it; it is required to draw a straight line from C, at right angles to AB. Take any point D in AC, and make CE equal to CD; from D, with any radius greater than DC, describe the arc FH, and from E with the F 1 OF GEOMETRY. BOOK I. 17 same radius, describe the arc FG, intersecting FH in F, and join CF; CF is perpendicular to AB. Join DF, EF. Because the triangles DFC, EFC, have the sides DF, CD equal to EF, CE, and the third side CF common; therefore (7. 1.), the angles CDF, CEF are equal; and therefore (4. 1.), the two triangles DFC, EFC are equivalent; and the angle DCF is equal to the angle BCF, and therefore (Def. 27.) CF is perpendicular to DE, or AB. Wherefore, from the given point C, in the given straight line AB, CF has been drawn at right angles to AB. Which was to be done. PROP. X. PROB. To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it. Let AB be the given straight line, which may be produced to any length both ways, and let C be a point without it. It is required to draw a straight line perpendicular to AB from the point C. Take any point D, upon the other side of AB, and from C as a centre (Def. 50.), with a radius CD, describe the arc FDG, meeting AB in FG; and bisect (8. 1.) FG in H, and join CF, CH, CG; the straight line C equal to the two sides GH, CH, each to each; and the base CF is equal (Def. 54.) to the base CG; therefore, the angle CHF is equal (7. 1.) to the angle CHG, and they are adjacent angles; but when a straight line, standing on a straight line, makes the adjacent angles equal to one another, each of them is a right angle, and the straight line which stands upon the other, is called a perpendicular to it (Def. 27.); therefore, from the given point C, a perpendicular CH has been drawn to the given straight line AB. Which was to be done. PROP. XI. THEOR. From a point without a straight line, only one perpendicular can be drawn to the line. Let DE be a straight line, A a point without it, and AB perpendicular to DE: no other perpendicular can be drawn from A to DE. If possible, let AC be also perpendicular. Produce AB to F, make BF equal to AB, and join FC. Because the angle ABC is a right angle, the angle FBC is also a right angle (Cor. 3. 1.) and the two triangles ABC, FBC have the side AB equal to BF, BC common, and the contained angle ABC, equal to the contained angle FBC; therefore, the triangles are equivalent (4. 1.), and the angle ACB is equal to the angle FCB; but ACB is a right angle by the hypothesis, therefore FCB is also a right angle; but when at a point in a straight line two other straight lines make the adjacent angles equal to two right angles, these two straight lines shall be in the same straight line (2. 1.); therefore ACF is a straight line; and between the points A and F there are two straight lines which do not coincide; but (Def. 21.) that is impossible; and, therefore, ACB is not a right angle. Therefore, From, &c. Q. E. D. PROP. XII. THEOR. If from the ends of the side of a triangle there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle. Let the two straight lines BD, CD be drawn from B, C, the ends of the side BC of the triangle ABC, to the point D within it. BD and DC are less than the other two sides BA, AC of the triangle. A Produce BD to E; and because two sides of a triangle are greater than the third side, the two sides BA, AE of the triangle ABE are greater than BE: to each of these add EC, therefore the sides BA, AC are greater than BE, EC: again, because the two sides CE, ED of the triangle CED are greater than CD, add DB to each of these; therefore the sides CE, EB are greater than CD, DB: but it has been shewn that BA, AC are greater than BE, B EC; much more then are BA, AC greater E D C than BD, DC. Therefore, If from the ends, &c. Q.E.D. |