404 



GEOMETRY. 



these two triangles will be identical. If the triangle 

 ABC, Jig. 42, have the side AC equal to the side BC; 

 then will the angle B be equal to the angle A. The 

 line which bisects the vertical angle of an isosceles 

 triangle, bisects the base, and is also perpendicular 

 to it. Every equilateral triangle, is also equiangular, 

 or has all its angles equal. If the triangle ABC, fig. 

 42, have the angle A equal to the angle B, it will 

 also have the side AC equal to the side BC. Every 

 equiangular triangle is also equilateral. Let the two 

 triangles ABC, ABD, fig. 43, have then 1 three sides 

 respectively equal, viz., the side AB equal to AB, 

 AC to AD, and BC to BD ; then sliall the two 

 triangles be identical, or have their angles equal, viz., 

 those angles that are opposite to the equal sides ; 

 namely, the angle BAG to the angle BAD, the angle 

 ABC to the angle ABD, and the angle C to the 

 angle D. 



45 



46 



Let ABC, fig. 44, be a triangle, having the side 

 AB produced to D ; then will the outward angle 

 CBD be greater than either of the inward opposite 

 angles A or C. Let ABC, fig. 45, be a triangle, 

 having the side AB greater than the side AC ; then 

 will the angle ACB, opposite the greater side AB, 

 be greater than the angle B, opposite the less side 

 AC. Let ABC, fig. 45, be a triangle ; then will the 

 sum of any two of its sides be greater than the third 

 side ; as, for instance, AC + CB greater than AB. 

 Let ABC, fig. 46, be a triangle ; then will the differ- 

 ence of any two sides, as AB AC, be less than 

 the third side BC. 



48 



Let the side AB, fig. 47, of the triangle ABC, be 

 produced to D ; then will the outward angle CBD 

 be equal to the sum of the two inward opposite 

 angles A and C. Let ABC, fig. 46, be any plane 

 triangle ; then the sum of the three angles A-f-B-f C 

 is equal to two right angles. If two angles in one 

 triangle, be equal to two angles in another triangle, 

 the third angles will also be equal, and the two 

 triangles equiangular. If one angle in one triangle 

 be equal to one angle in another, the sums of the 

 remaining angles will also be equal. If one angle of 

 a triangle be right, the sum of the other two will also 

 be equal to a right angle, and each of them singly 

 will be acute, or less than a right angle. The two 

 least angles of every triangle are acute, or each less 

 than a right angle. Let ABC, fig. 48, be a right- 

 angled triangle, having the right angle at C ; then 

 will the square of the hypotenuse AB, be equal to 

 the sum of the squares of the other two sides AC, 

 CB. Or AB = AC* -f BC*. 



49 50 '51 



Let ABC, fig. 49, be any triangle, having CD per- 

 pendicular to AB ; then will the difference of the 

 squares of AC, BC, be equal to the difference of the 

 squares of AD, BD ; that is, AC 1 BC^AD" BD. 

 Let ABC, fig. 49, 2nd, be a triangle, obtuse-angled at 

 A, and CD perpendicular to AB; then will the square 

 of AC be greater than the squares of AB, BC, by 

 twice the rectangle of AB, BD. That is, AC*= AB S + 

 BC" + 2AB. BD. Let ABC, fig. 49, 1st, be a triangle, 

 having the angle A acute, and CD perpendicular to 

 AB; then will the square of BC, be less than the 

 squares of AB, AC, by twice the rectangle of AB, 

 AD. That is, BC" = AB* + AC" 2AD . AB. 

 Let ABC, fig. 50, be a triangle, and CD the line 

 drawn from the vertex to the middle of the base AB, 

 bisecting it into the two equal parts AD, DB ; then 

 will the sum of the squares of AC, CB, be equal to 

 twice the sum of the square of CD, AD; or AC* + 

 CB" = 2CD* + 2AD*. Let ABC, fig. 51, be an isos- 

 celes triangle, and CD a line drawn from the vertex 

 to any point D in the base : then will the square of 

 AC, be equal to the square of CD, together with the 

 rectangle of AD and DB. That is, AC" = CD 8 + 

 AD . DB. 



Let ABC, DBF, fig. 52, be two triangles, having 

 the angle A = the angle D, and the sides AB, AC, 

 proportional to the sides DE, J)F ; then will the 

 triangle ABC be equiangular with the triangle DEF. 

 Let ABC, fig. 53, be a right-angled triangle, and CD 

 a perpendicular from the right angle C to the hypo- 

 tenuse AB ; then will CD be a mean proportional 

 between AD and DB; AC a mean proportional be- 

 tween AB and AD ; BC a mean proportional between 

 AB and BD. Let the two triangles ADC, DEF, fig. 

 54, have the same altitude, or be between the same 

 parallels AE, CE ; then is the surface of the triangle 

 ADC, to the surface of the triangle DEF, as the 

 base AD is to the base DE. Or, AD : DE : : the 

 triangle ADC : the triangle DEF. 



55 56 57 



A G B II E 



Let ABC, BEF, fig. 55, be two triangles having 

 the equal bases AB, BE, and whose altitudes are the 

 perpendiculars CG, FH; then will the triangle ABC : 

 the triangle BEF : : CG : FH. Triangles when 

 their bases are equal, are to each other as their alti- 

 tudes ; and when their altitudes are equal, they are 

 to each other as their bases ; therefore, universally, 

 when neither are equal, they are to each other in 

 the compound ratio, or as the rectangle or product of 

 their bases and altitudes. Let DE, fig. 56, be par- 

 allel to the side BC of the triangle ABC, then will 

 AD : DB : : AE : EC. 



AB : AC : : AD : AE, 



AB : AC : : BD : CE. 



Let the angle ACB, fig. 57, of the triangle ABC, be 

 bisected by the line CD, making the angle r equal 

 to the angle s : then will the segment A D be to the 

 segment DB,as the side AC is to the side CB. Or 

 AD :DB :: AC :CB 



