If two triangles have two sides of the one respectively equal to two sides of the other, and also have their bases equal, the angles opposite the equal bases are equal, and the two triangle. are equal in all respects. Let ABC and DEF be two triangles, having the side AB equal to DE, the side B C to EF, and the base AC equal to the base DF; then the angle в is equal to the angle E; and the two triangles are equal in all respects. A B E For if the triangle ABC be so applied to the triangle D E F, that the point a may be on D, and the base AC upon D F, then the point c must coincide with F, because AC and DF are equal (Hyp.); and as the base AC coincides with DF, and the sides AB and BC are equal to the sides DE and EF (Hyp.), the vertex B must fall on E (Prop. 7.), and the equal sides must coincide (Ax. 10.); therefore the angles в and E must also coincide, and are therefore equal (Ax. 8.). And as the three sides of one triangle coincide with the three sides of the other, the triangles themselves must also coincide, and are therefore equal (Ax. 8.). Which was to be demonstrated. COR. Two triangles are equal, if they have a common side, and the remaining sides of the one equal to the remaining sides of the other. PROP. IX. PROB. To bisect a given rectilineal angle. Let BAC be the given angle, it is required to bisect it. D E Take any point D in the side a B, and from AC cut off a part A E equal to A D (Prop. 3.): join DE, and upon it describe an equilateral triangle DEF (Prop. 1.) at the side opposite to A; join A F; then the angle BAC or DAE is bisected by the line a F. For in the triangles DAF and EAF, the side DA is equal to E A (Const.), and a F is common to both, and the base D F is equal to E F (Const.); therefore the angle DAF is equal to EAF (Prop. 8.), and therefore the given angle BAC is bisected by the line AF. Which was to be done. B F C PROP. X. PROB. To bisect a given right line. Let A B be the given line, it is required to bisect it. Upon A B describe an equilateral triangle ABC (Prop. I.), and bisect the angle ACB by the line CD (Prop. 9.); then AB is bisected in the point D. A D B For in the triangles ACD and BCD, the side a c is equal to BC (Const.), the side CD is common to both, and the angle A CD is equal to BCD (Const.); therefore the base AD is equal to BD (Prop. 4.), and therefore the given line AB is bisected in D. Which was to be done. COR. By this proposition a line may also be divided into 4 equal parts, or into 8, 16, &c., by bisecting again each part. PROP. XI. PROB. To draw a right line perpendicular to a given line from a given point in it. Let AB be the given line, and c the given point; it is required to draw from c a line which shall be perpendicular to the line AB. D C Take any point D in AC, and cut off from CB a part CF equal to CD (Prop.3.), A and upon DF describe an equilateral triangle DFG (Prop. 1.); join &c; then the line GC is dicular to A B. F B perpen For in the triangles CDG and CFG, the sides CD and DG in the one, are equal to the sides cr and FG in the other (Const.), and the side CG is common to both, therefore the angles DCG and FCG opposite the equal sides are equal (Prop. 8.), and these are the adjacent angles; therefore the line ac is perpendicular to AB (Def. 11.), and it is drawn from the point c. Which was to be done. COR. In like manner a perpendicular can be drawn from the extremity of a given line, by first producing the line. B PROP. XII. PROB. To draw a right line perpendicular to a given indefinite line, from a given point not in it. Let AB be the given line, and c the given point; it is required to draw from c, a line which shall be perpendicular to AB. Take any point D, on the other side of A B, and from the centre c, with the radius CD, describe a circle cutting the line AB, A C or the line produced in the points E and F; bisect EF in the point & (Prop. 10.), and join ca; then the line CG is perpendicular to A B. For draw the lines CE and CF; then in the triangles GEC and GFC, the side CE is equal to CF, being radii of the same circle, and EG is equal to FG (Const.), and the side CG is common to both; therefore the angles CGE and CGF opposite the equal sides are equal (Prop. 8.), and these are the adjacent angles; therefore CG is perpendicular to AB (Def. 11.), and it is drawn from the point c. Which was to be done. PROP. XIII. THEOR. The angles made by one line standing upon another are together equal to two right angles. Let the line AB stand upon CD, making with it the angles ABC and ABD; those angles are together equal to two right angles. A D C D B B First. If the line AB be perpendicular to CD, then each of the angles ABC and ABD is a right angle (Def. 16.); therefore both together are two right angles. Secondly. If the line AB be not perpendicular to CD, from the point B draw BE perpendicular to CD (Prop. 11.); then it is evident that the angles ABC and ABD together are equal to the angles EBC and EBD: but EBC and EBD are two right angles (Const.), therefore ABC and A BD together are equal to two right angles. Which was to be demonstrated. COR. 1. The angles made by several lines standing upon the same right line, at the same point, are together equal to two right angles. COR. 2. The angles made by any number of lines meeting at a point are together equal to four right angles. If two right lines meet another right line, at the same point, at opposite sides, and make angles with it equal to two right angles, the two lines form one continued right line. Let the lines AB and BC meet the line DB at the point в, and make with it the angles DBA and DBC equal to two right angles ; then a B and B C. form one right line. A B D E C For, if AB and BC do not form one right line, let any other line BE form with AB a continued right line; then, because AE is a right line and DB stands upon it, the angles DBA and DBE are equal to two right angles (Prop. 13.); but the angles DBA and DBC are also equal to two right angles (Hyp.), therefore the angles DBA and DBC are equal to DBA and DBE (AX. 1.). Take away the angle DBA, which is common to both, and the remaining angles DBE and DBC are equal (Ax. 3.), a part equal to the whole which is absurd; therefore BE does not form a right line with AB; and in the same manner it can be proved that no other line but BC can form a right line with AB: therefore A B and BC form one continued right line. W. W. D. PROP. XV. THEOR. If two right lines cut one another, the vertical or opposite angles are equal. Let the lines A B and C D cut one another at E; then the angle DEB is equal to CEA, and the angle DEA is equal to BEC. For, as the line DE stands upon в A, the angles DEB and DEA are equal to two right angles (Prop. 13.), and as the line A E E B stands upon DC, the angles DEA and CEA are likewise equal to two right angles; therefore the angles DEB and DEA are equal to the angles CEA and DEA (AX. 1.). Take away the angle DEA, which is common to both, and the remaining angles DEB and CEA are equal (Ax. 3.), and these are the vertical angles. In the same manner it can be proved that the angles DE A and BEC are equal. Which was to be demonstrated. PROP. XVI. THEOR. If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. Let the side AB of the triangle ABC be produced to D; the exterior angle CBD is greater than either of the interior opposite angles CAB or ACB. Bisect the side вс in E (Prop. 10.); draw AE and produce it till EF is equal to AE (Prop. 3.); join FB. A P E B D Then in the triangles ACE and BEF the sides A E and EC in the one, are equal to the sides FE and EB in the other (Const.), and the vertical angles AEC and BEF are equal (Prop. 15.); therefore the angles EBF and ACE are also equal (Prop. 4.) ; but the angle CBD is greater than CBF, therefore CBD is greater than ACE or AC B. In like manner, if the side CB be produced and AB be bisected, it can be shown that the angle ABG is greater than the interior angle CAB; but ABG is equal to CBD, being vertical angles (Prop. 15.); therefore CBD is also greater Which was to be demonstrated. than CAB. PROP. XVII. THEOR. Any two angles of a triangle are together less than two right angles. Let ABC be a triangle; any two of its angles are together less than two right angles. Produce the side AC to D (Cor. 2.), then the exterior angle BCD is greater than the interior angle B A C (Prop. 16.); B C D add to each the angle BCA; and the angles BAC and BCA are less than the angles BCD and BCA; but BCD and BCA are equal to two right angles (Prop. 13): therefore BAC and BCA are less than two right angles. In like manner, by producing the other sides, it can be shown that the angles BAC and ABC, or the angles ACB and ABC are less than two right angles. Which was to be demonstrated. COR. 1. If one angle of a triangle be obtuse or right, the other two angles must be acute. COR. 2. The angles at the base of an isosceles triangle are both acute. |