GEOMETRY. 



case of this proposition, the triangles are equal in 

 every respect. 



(4.) Let ABC, DEF be two triangles, having 

 the two angles B and C in the one triangle equal 

 respectively to the two E and F in the other, and 

 the side AB in the one equal to the corresponding 

 side DE in the other ; it is required to prove that 



the two triangles are equal in every respect. For if 

 ABC be applied to DEF, so that A lies on D, and 

 AB along DE, then B shall lie on E, because AB 

 is equal to DE, and BC shall lie on EF, because 

 the angle ABC is equal to the angle DEF, and C 

 shall lie on F ; for, if it do not, let it lie at G in 

 EF produced, then GD will be the position of AC, 

 and DGE will represent the angle ACB, and be 

 equal to angle DFE ; but DFE is greater than 

 DGE, since it is the exterior angle, therefore the 

 point C must lie on F, and hence the triangles must 

 be equal in every respect. 



(5.) Let ABC, DEF be two triangles, having 

 the two sides AB, AC in the one triangle equal re- 

 spectively to the two sides DE, DF in the other, 

 and the angle B in the one equal to the corre- 

 sponding angle E in the other, the remaining 

 angles C and F being each greater or each less 

 than a right angle ; it is required to prove that 

 the triangles are equal in every respect For if 



the triangle ABC be applied to DEF, so that A 

 may lie on D, and AB on DE, then B shall lie on 

 E, for AB is equal to DE ; and angle B being 

 equal to angle E, BC shall lie along EF, and C 

 shall lie on F ; for if it does not, it can only lie at 

 one other point G, in EF, since only two equal 

 lines can be drawn from D to the line EF. But 

 the point C cannot lie at G, since the two angles 

 ACB, DFE are supposed to be both acute or 

 both obtuse ; hence the point C must lie on F, 

 and therefore the two triangles are equal in every 

 respect 



CONSTRUCTIONS (l.). 



(i.) To describe a triangle of which the three 

 sides are equal to three given lines. 



Let A, B, and C be the three given lines. Take 

 DE equal to A, and 

 from D as centre, with 

 radius equal to B, de- 

 scribe a circle ; and 

 from E as centre, with 

 radius equal to C, de- 

 scribe a circle. Let / 

 the two circles cut \ 

 each other in F and G. 

 Join DF, EF; then 

 DFE will evidently be 

 the triangle required, so also will DGE. The 

 problem is evidently impossible (Prop. VI.), when 



any two of the sides together are not greater than 

 the third side. It follows that an isosceles triangle 

 may be described upon a given line. Also that 

 an equilateral triangle may be described upon a 

 given line. 



(2.) To bisect a given line. 

 Let AB be a given line ; it is required to bisect 

 it. From A and B as centres, and with the same 

 length as radius, describe two 

 circles which cut each other 

 in C and D. Join CD, and 

 let this line cut AB in E. E 

 is the middle of AB. Join 

 AC, BC, AD, and BD. Be- 

 cause, in the two triangles 

 ACD and BCD, we have AC 

 in the one equal to BC in the 

 other, AD in the one equal to 

 BD in the other, and CD common to both tri- 

 angles ; therefore (Prop. VII. 3) the angle ACE 

 is equal to the angle BCE. Again, in the two 

 triangles ACE, BCE, we have the side AC equal 

 to BC and CE common, and the contained angle 

 ACE in the one equal to the contained angle BCE 

 in the other ; therefore (Prop. VII. i) the side AE 

 is equal to the side BE, and AB is bisected in E. 

 Cor. Hence a line may be divided into four, 

 eight, sixteen, &c. equal parts. 

 (3.) To bisect a given angle. 

 Let ABC be the given angle which it is required 

 to bisect From B as centre, 

 describe an arc cutting BA in 

 D, and BC in E. From D and 

 E as centres, and with the same 

 length as radius, describe two 

 arcs cutting in F. The line which 



joins B and F bisects the angle 



at B. For, in the two triangles 



BDF, BEF, we have the three 



sides of the one equal to the 



three sides of the other, and 



(Prop. VII. 3) the angle DBF is equal to the angle 



EFB ; hence BF bisects the angle at B. 



Cor. i. Hence an angle may be divided into 

 four, eight, sixteen, &c. equal parts. 



Cor. 2. The right angle may be divided into 



three equal parts. 



Let CAB be a right angle; it is required to 



divide it into three equal parts. 

 On AB, one of its sides, describe c 



the equilateral triangle ADB ; 



then each angle of this triangle 



is one-third of two right angles 



(Prop. III. 3), or two-thirds of 



one right angle, therefore the A 



angle CAD is one-third of a 



right angle. Let the angle DAB be bisected by 



the line AE ; then the right angle CAB is divided 



into three equal parts by the two lines AD, AE. 



(4.) From a given point, to draw a line perpen- 

 dicular to a given line. 



Let C be the given point, and AB the given 



line ; it is required to 



draw from the point C 



a line perpendicular to 



AB. 



First, If the point C be 



situated on the given line. ^ 



This is the same as bi- \ 



secting the angle ACB, 



and the construction of die preceding proposition 



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