CHAMBERS'S INFORMATION FOR THE PEOPLE. 



\ 



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 'F\ 



will give the required line. For from C as centre, 

 with any radius, describe an arc cutting AB in D 

 and E. Then from D and E as centres, with 

 the same length as radius, describe two arcs 

 cutting in F. Join FC. Then FC bisects the 

 angle ACB, or it is perpendicular to AB. 



Second, If the point C be situated not on the 

 given line. From C as 

 centre, describe a circle 

 cutting AB in D and E. 

 Then from D and E as 

 centres, with the same 

 length as radius, describe 

 two arcs cutting in F. 

 Join FC. FC is perpen- 

 dicular to AB ; for it is 

 evident that the angle 

 CGD is equal to the angle CGE, and these are 

 adjacent angles ; therefore each is a right angle, 

 and CG is perpendicular to AB. 



Cor. It is evident that only one perpendicular 

 can be drawn from a given point to a given line, 

 whether the point is on the line or without it. 



(5.) At a given point in a given line, to make an 

 angle equal to a given angle. 



Let A be the given point, AB the given line, 

 and C the given angle ; it is required to make an 

 angle at A equal 

 to the given angle 

 C. In CD and 

 CE, the lines con- 

 taining the given 

 angle take any 

 points D and E. 

 Join DE. From 

 AB cut off AF 

 equal to CD. From A as centre, with radius 

 equal to CE, describe an arc ; from F as centre, 

 with radius equal to DE, describe an arc. Let 

 these arcs meet in G. Join AG. Then the angle 

 which has been thus formed at A, is equal to the 

 angle at C. 



PROP. VIII. If two triangles have two sides 

 of the one equal to two sides of the other, but the 

 contained angles unequal, the base of that which 

 has the greater angle is greater than the base of 

 the other. 



Let ABC, DEF be two triangles, having AB in 

 the one equal to DE in the other, and AC in the 

 one equal to DF in the other, but the angle BAG 

 greater than the angle EOF ; it is required to 

 prove that the base BC is greater than the base 

 EF ; for if the triangle ABC be lifted up, and 

 applied to DEF, so that A lies on D, and AC on 

 DF, then C shall lie on F, for AC is equal to DF. 

 Let DG be the position of AB, then BC shall 

 take the position FG, and it is required to prove 

 that FG is greater than EF. Bisect the angle 

 EDG by the line DH. Join EH. Then in the 

 two triangles EDH, GDH, we have two sides and 



the contained angle in the one equal to two sides 

 and the contained angle in the other, therefore 

 the base EH of the one is equal to GH of the other. 



614 



Add FH to each of these, then we have EH and 

 HF together equal to GF ; but EH and HF are 

 greater than EF, therefore FG is greater than 

 EF ; but FG is the same as BC, therefore BC is 

 greater than EF. 



Conversely, if two triangles have two sides oj 

 the one equal to two sides of the other, but their 

 bases unequal, the angle contained by the two sides 

 of that which has the greater base is greater than 

 the angle contained by the two sides of the other. 

 This is proved by supposing it <?/true, and shew- 

 ing that a contradiction follows. 



PARALLEL LINES. 



PROP. IX. Two straight lines are parallel 

 when a line intersecting them makes : 



(i.) The alternate angles equal. 



(2.) The exterior angle equal to the interior,, 

 and remote on the same side. 



(3.) The two interior angles on the same side 

 equal to two right angles. 



(i.) Let AB and CD be two lines intersected by 

 a third line EG, making the alternate angle AFG 

 equal to the alternate 

 angle FGD ; it is re- 

 quired to prove that 

 AB is parallel to CD. 

 If AB and CD be not 

 parallel, let them meet 

 in H. Then HFG will 

 be a triangle, and the 



exterior angle AFG will be greater than the in- 

 terior remote angle FGD ; but by hypothesis 

 these angles are equal. They are therefore equal 

 and unequal, which is absurd ; therefore AB and 

 CD do not meet that is, they are parallel 



(2.) Let AB and CD be two lines intersected by 

 EF, making the exterior angle EFB equal to the 

 interior angle FGD ; it 

 is required to prove 



that AB is parallel to 



CD. If AB and CD ^ ~^=~" 



be not parallel, let them c /' 5 



meet in H. Then HFG / 



will be a triangle, and 



the exterior angle EFB will be greater than the 



interior angle FGD ; but by hypothesis these 



angles are equal. They are therefore equal and 



unequal, which is absurd ; therefore AB and CD 



do not meet that is, they are parallel 



(3.) Let AB and CD be two lines, and EG 

 another line intersecting them, and making the 

 two angles BFG, FGD together equal to two 

 right angles ; it is required to prove that AB and 

 CD are parallel. If AB 

 and CD be not parallel, 

 let them meet in H. 

 Then HFG will be a 

 triangle, and any two 

 of its angles HFG, 

 FGH will be together 

 less than two right 



angles ; but by hypothesis they are together equal 

 to two right angles. They are therefore together 

 equal to two right angles, and together less than 

 two right angles, which is absurd ; therefore AB 

 and CD do not meet if produced that is, they 

 are parallel. 



Conversely, if a line intersect two parallel lines : 



(i.) The alternate angles are equal. 



