CHAMBERS'S INFORMATION FOR THE PEOPLE. 



CD, and between the same parallels AD and HE. 

 Join AF, BE. Then 



the figure ABEF is H c r E_ 



a parallelogram, since 

 its opposite sides AB 



and FE are both 



equal and parallel A c 



(Prop. X.); and the 



parallelogram ABEF is equal to the parallelogram 



ABGH, since they stand on the same base and 



between the same parallels. It is also equal to 



CDEF ; therefore ABGH is equal to CDEF. 



Cor. I. Triangles upon the same base, and 

 between the same parallels, 

 are equal 



Let ABC and ABD be two 

 triangles on the same base 

 AB, and between the same 

 parallels AB and CD ; then 

 they are equal, being halves of equal parallel- 

 ograms. 



Cor. 2. Triangles upon equal bases, and be- 

 tween the same parallels, c f 

 are equal. 



Let ABC, DEF be two 

 triangles on the equal 

 bases AB, DE, and be- 

 tween the same parallels 

 AE and CF ; then they are equal, being halves of 

 equal parallelograms. 



Conversely, equal triangles upon the same side 

 of equal bases are between the same parallels. 



Let ABC, ABD be equal triangles upon the 

 same side of the same base AB, then they are 

 between the same parallels 

 that is, DC is parallel to AB. 

 If DC be not parallel to AB, 

 draw DE parallel. Join EB. 

 Then the two triangles ADB, 

 AEB are equal. But ADB 

 is equal to ACB, therefore 

 AEB is equal to ACB, which is absurd ; there- 

 fore DE cannot be parallel to AB, and no other 

 line than DC through D can be. 



Cor. 3. If a parallelogram and triangle be upon 

 the same base, and between the same parallels, 

 the parallelogram is double of the triangle. 



Let ABC be a triangle, and ABDE a paral- 

 lelogram on the same base AB, and between the 

 same parallels. The paral- 

 lelogram is double of the 

 triangle. Join BE. Then 

 the two triangles ABE, 

 ABC are equal, being on 

 the same base, and between 



the same parallels ; but ABDE is double of 

 ABE, therefore ABDE is also double of the 

 triangle ABC. 



CONSTRUCTIONS (ll.). 



(i.) Through a given point to draw a line 

 parallel to a given line. 



Let AB be the given line, C the given point ; it 

 is required to draw through C 



a line parallel to AB. Take - ? 



any point, D in AB, and join / 



CD. At the point C, in the / 



line CD, make the angle ^ D 



ECD equal to the angle 



CDB : then EC will be parallel to AB, since these 



616 



are two lines intersected by a third, CD, making 

 the alternate angles equal. 



(2.) To transform a triangle into a parallelogram 

 having an angle equal to a given angle. 



Let ABC be the given triangle, which it is 

 required to transform into a parallelogram, having 

 one of its angles equal to a given angle. Bisect 

 the base BC in D. Join AD. Then the whole 

 triangle ABC is double of 

 the triangle ADC. At the 

 point D in DC make the 

 angle CDE equal to the 

 given angle, and through 

 C draw CF parallel to B o c 

 ED ; through A draw 



AEF parallel to BC, then DCFE is a parallel- 

 ogram, and it is double of the triangle ADC 

 (Prop. XI. Cor. 3) ; but ABC is also double of this 

 triangle, therefore EDCF is equal to ABC, and 

 has an angle equal to a given angle. 



THE CIRCLE. 



The Chord and Tangent 



PROP. XII. The line which bisects a chord at 

 right angles passes through the centre of the circle. 



Let AB be any chord, having C for its middle 

 point. If a line CQ be drawn through C perpen- 

 dicular to AB, it passes through the centre. For 

 if the centre be not in this 

 line, let it be any point P. 

 Join PA, PC, PB. Then in 

 the two triangles PAC, PBC 

 we have PA equal to PB, AC 

 equal to CB, and PC com- 

 mon ; therefore (Prop. VII. 3) 

 the angle PCA is equal to the 

 angle PCB, therefore each is 

 a right angle. But QCB is a right angle, there- 

 fore QCB is equal to PCB, which is absurd ; 

 therefore the centre of the circle lies on the line 

 QC. 



Hence to find the centre 

 of the circle. Draw any two 

 chords AB, CD, which are 

 not parallel. Bisect each of 

 these chords, and through 

 the points of bisection, E and 

 F, draw lines at right angles 

 to AB and CD ; these lines 

 will each pass through the 

 centre, and therefore the centre will be their point 

 of intersection. 



Conversely, (i.) A line drawn from the centre 

 perpendicular to a chord bisects it. 



(2.) A line drawn from the centre to the middle 

 of a chord is perpendicular to the chord. 



(i.) If the line OC, drawn from the centre O 

 perpendicular to the chord 

 AB, meets it in C ; it is 

 required to prove that C is 

 the middle of AB. In the 

 two triangles AGO, BCO we 

 have the two angles AGO, 

 OAC (Prop. IV.) in the one 

 equal to the two BCO, OBC 

 in the other, and the side OC 

 is common to both ; therefore 

 their other sides are equal namely, AC to CB ; 

 therefore C is the middle of AB. 



