Diagram 46.
Diagram 46.
Diagram 46.
We wish to prove, that,
If a straight line intersects two other straight lines so that two interior angles on the same side of the intersecting line are equal to two right angles, the two lines are parallel.
Let the straight linee fintersect the two straight linesa b,c d, in the pointsgandh, so that the angles Red, Blue, are equal to two right angles.
Then will the linesa b,c d, be parallel.
For the angles Red, Blue, are supposed equal to two right angles.
The adjacent angles Red, Green, are known to be also equal to two right angles.
Then the interior angles Red, Blue, are equal to the adjacent angles Red, Green.
If from the interior angles Red, Blue, we take away the angle Red, we have left the angle Blue.
If from the adjacent angles Red, Green, we take the same angle Red, we shall have left the angle Green.
Then the angle Blue is equal to the angle Green.
But the angle Blue measures the direction of the lineh dfrom the linee f.
And the angle Green measures the direction of the lineg bfrom the linee f.
Then the linesg b,h d, have the same direction, and are parallel.
1. Prove the same without the colors.
2. Prove the same, using the anglef h d.
3. Prove the same, supposing the anglesa g h,g h c, equal to two right angles, and using the anglea g e.
4. Prove the same, using the anglec h f.
See NoteE, Appendix.
PROPOSITION VIII. THEOREM.
The following demonstration is very easy. Read it once, and see if you can go through it without a second reading:—
We wish to prove that
The sum of any two sides of a triangle is greater than the third side.
Let the figurea b cbe a triangle, then will the sum of any two sides, asa c,c b, be greater than the third sidea b.
For the straight linea bis the shortest distance between the two pointsaandb, and is therefore less than the broken linea c b.
The following solution is so easy that you will understand it at once:—
We wish
To construct an equilateral triangle on a given straight line.
Leta bbe the given line.
With the pointaas a centre, anda bas a radius, draw the circumference of the circle, or a part of one.
With the pointbas a centre, and the same radiusa b, draw another circumference, or a part of one.
From the pointc, in which the circumferences or arcs intersect, draw the straight linesa candb c.
Now, because the linesa banda care radii of the same circle, they are equal.
And, because the linesa bandb care radii of the same circle, they are also equal.
Then, because the two linesa c,b c, are separately equal to the linea b, they are equal to each other, and the triangle is equilateral.
Let the figurea b cbe a triangle.
Produce the sidea ctod.
We have now another angle,b c d, and we wish to find out if it is equal to any of the angles of the triangle.
From the pointcdraw the linec eparallel toa b.
Because the straight linea dintersects the two parallelsa b,c e, the angleais equal to what other angle?
Because the straight lineb cintersects the two parallelsa b,c e, the anglebis equal to what other angle?
Then the anglesaandbare equal to what two angles?
How does the angleb c dcompare with the anglesb c e,e c d?
Then, if the anglesaandb, on the one hand, and the angleb c d, on the other, are separately equal to the anglesb c e,e c d,
What have you found out?
What axiom have you just employed?
To what same thing have you found two other things equal?
What two things did you find equal to it?
We wish to prove, that,
If any side of a triangle be produced, the new angle formed will be equal to the sum of the angles that are not adjacent to it.
Leta b cbe a triangle.
Produce the sidea ctod; then will the new angleb c dbe equal to the sum of the anglesaandb.
For from the pointcdrawc eparallel toa b.
Then, because the straight linea dintersects the two parallelsa b,c e, in the pointsaandc,
The opposite exterior and interior anglesaande c dare equal to each other.
And because the straight lineb cintersects the same parallels in the pointsbandc,
The interior alternate anglesbandb c eare equal.
Then the anglesaandbof the triangle are equal to the anglesb c eande c d.
But the new angleb c dis equal to the anglesb c e,e c d.
Then because the new angleb c d, and the anglesaandbare separately equal to the anglesb c e,e c d, they are equal to each other.
Let the figurea b cbe a triangle.
Produce the sidea ctod.
By the last theorem, the angleb c dis equal to what angles of the triangle?
What angle must we add to these angles to make up the three angles of the triangle?
If we add the same angle to the angleb c d, what adjacent angles do we get?
Then the three angles of the triangle,a,b, andc, are equal to what two angles?
But the adjacent anglesa c bandb c dare equal to what?
Then, because the three angles of the triangle,a,b, andc, and two right angles, are separately equal to the two adjacent anglescandb c d.
What new thing have you found out?
DEMONSTRATION.
We wish to prove that
The three angles of any triangle are equal to two right angles.
Let the figurea b cbe a triangle; then will the sum of the anglesa,b, andc, be equal to two right angles.
For, produce the sidea ctod.
The new angleb c dis equal to the sum of the anglesaandb.
If to the anglesaandbwe add the anglec, we shall have the three angles of the triangle.
If to the angleb c dwe add the same anglec, we shall have the adjacent anglescandb c d.
Then the three angles of the trianglea,b,c, are equal to the adjacent anglescandb c d.
But the adjacent anglescandb c dare equal to two right angles.
Then, because the three angles of the triangle are equal to the adjacent anglescandb c d, they are equal to two right angles.
Let the Fig. A B C D be a parallelogram.
Produce the side C D to F.
Because the straight line B D intersects the parallels A B and C F, the angle B is equal to what other angle?
Because the straight line C F intersects the parallels A C and B D, the angle C is equal to what other angle?
Then what follows from this?
To what angle did you find two others equal?
What two angles did you find equal to it?
What axiom do you think of?
See if you can go through the demonstration without reading it even once.
We wish to prove that
The opposite angles of a parallelogram are equal to each other.
Let the Fig. A B C D be a parallelogram.
Then will any two opposite angles, as B and C, be equal to each other.
For produce the line C D to F.
Because the straight line B D meets the two parallels A B and C F,
The interior alternate angles B and E are equal to each other.
Because the straight line C F meets the two parallels B D and A C,
The opposite exterior and interior angles C and E are equal to each other.
Then, because the angles B and C are separately equal to the angle E, they are equal to each other.
1. Prove the same by producing the line A B towards the left.
2. Prove the same by producing the line B D downwards.
3. Prove the angles A and D equal to each other by producing the line C D towards the left.
4. Prove the same by producing the line D B upwards.
5. See if you can prove the same by drawing a diagonal through the points A and D.
In these two triangles we have tried to make the sidea bof the one equal to the sided eof the other; the sidea cof the one equal to the sided fof the other; and the included angleb a cof the one equal to the included anglee d fof the other.
We now wish to find out if the third sideb cof the one is equal to the third sidee fof the other, and if the two remaining anglesbandcof the one are equal to the two remaining angleseandfof the other.
Suppose we were to cut the triangled e fout of the page, and place it upon the trianglea b c, so that the lined eshould fall upon the linea b, and the pointdupon the pointa.
As the lined eis equal to the linea b, upon what point will the pointefall?
If the anglee d fwere less than the angleb a c,would the lined ffall within or without the triangle?
If the anglee d fwere greater than the angleb a c, where would the lined ffall?
Since the angleais equal tod, where, then, must the lined ffall?
As the lined fis equal to the linea c, upon what point will the pointffall?
Then, if the pointefalls upon the pointb, and the pointfupon the pointc, where will the linee ffall?
Now, because the three sides of the triangled e fexactly fall upon the three sides of the trianglea b c, we saythe two magnitudes coincide throughout their whole extent, and are therefore equal.
What three parts of the trianglea b cdid we suppose to be equal to three corresponding parts of the triangled e fbefore we placed one upon the other.
What line of the one do wefindequal to a line in the other?
What two angles of the one do wefindequal to two angles in the other?
What do you think of the areas of the triangles?
We wish to prove, that,
If two triangles have two sides, and the included angle of the one equal to two sides and the included angle of the other, each to each, the two triangles are equal in all respects.
Let the trianglesa b candd e fhave the sidea bof the one equal to the sided eof the other; the sidea cof the one equal to the sided fof the other; and the included angleb a cof the one equal to the included anglee d fof the other, each to each; then will the two triangles be equal in all their parts.
For, place the triangled e fupon the trianglea b c, so that the lined eshall fall upon the linea b, with the pointdupon the pointa.
Because the lined eis equal to the linea b, the pointewill fall upon the pointb.
Because the anglee d fis equal to the angleb a c, the lined fwill fall upon the linea c.
Because the lined fis equal to the linea c, the pointfwill fall upon the pointc.
Then, because the pointeis on the pointb, and the pointfon the pointc, the linee fwill coincide with the lineb c, and the two triangles will be found equal in all their parts;
That is, the angleeis found to be equal to the angleb, the anglefto the anglec, the linee fto the lineb c, and the area of the trianglea b cto the area of the triangled e f.
In these two triangles we have tried to make the anglebof the one equal to the angleeof the other; the anglecof the one equal to the anglefof the other; and the included sideb cof the one equal to the included sidee fof the other.
We now wish to find out if the remaining angleaof the one is equal to the remaining angledof the other, and if the two remaining sidesa banda cof the one are equal to the two remaining sidesd eandd fof the other.
Suppose we were to cut the triangled e fout of the page and place it upon the trianglea b c, so that the linee fshall fall upon the lineb c, with the pointeupon the pointb.
Because the linee fis equal to the lineb c, upon what point will the pointffall?
Because the angleeis equal to the angleb, where will the linee dfall?
Because the anglefis equal to the anglec, where will the lined ffall?
Then, if the lined efalls upon the linea band the lined fupon the linea c, where will the pointdfall?
Now because the three sides of the triangled e fexactly fall upon the three sides of the trianglea b c, we saythe two magnitudes coincide throughout their whole extent, and are therefore equal.
Suppose the angleewere greater than the angleb, would the linee dfall within or without the triangle?
If it were less, where would the line fall?
Why does the lined efall exactly upon the linea b?
We wish to prove that,
If two triangles have two angles, and the included side of the one equal to two angles and the included side of the other, each to each, the two triangles are equal to each other in all respects.
Let the trianglesa b candd e fhave the anglebof the one equal to the angleeof the other; the anglecof the one equal to the anglefof the other; and the included sideb cof the one equal to the included sidee fof the other, each to each; then will the two triangles be equal in all their parts.
For place the triangled e fupon the trianglea b c, so that the linee fshall fall upon the lineb c, with the pointeupon the pointb.
Because the linee fis equal to the lineb cthe pointfwill fall upon the pointc.
Because the angleeis equal to the angleb, the linee dwill fall upon the lineb a, and the pointdwill be somewhere in the lineb a.
Because the anglefis equal to the anglec, the linef dwill fall upon the linec a, and the pointdwill be somewhere in the linec a.
Then, because the pointdis in the two lines,b aandc a, it must be in their intersection, or upon the pointa.
And, as the two triangles coincide throughout their whole extent, they are equal in all their parts.
That is, the angleais found to be equal to the angled; the sideb ato the sidee d; the sidec ato the sidef d; and the area of the trianglea b cto the area of the triangled e f.
We wish to prove that
The opposite sides of any parallelogram are equal.
Let the figurea b c dbe a parallelogram; then will the sidesa bandc dbe equal to each other; likewise the sidesa dandb c.
For, draw the diagonalb d.
Because the figure is a parallelogram, the sidesa bandd care parallel, and the interior alternate anglesnandoare equal.
Because the figure is a parallelogram, the interior alternate anglesrandmare equal.
Then the two trianglesa d b,b d c, have two angles and the included side of the one equal to two angles and the included side of the other, each to each, and are therefore equal;
And the sidea bopposite the anglemis equal to the sidec dopposite the equal angler;
And the sidea dopposite the anglenis equal to the sideb copposite the equal angleo.
Prove the same by drawing a diagonal fromatoc.
SupposeA Bto be a straight line, andCany point out of it.
From the pointCdraw a perpendicularC FtoA B.
Let us see if this perpendicular is not shorter than any other line we can draw from the same point to the same line.
Draw any other line fromCtoA BasC E.
Now, asC Eis any line whatever other than a perpendicular, if we find that the perpendicularC Fis shorter than it we must conclude that it is the shortest line that can be drawn fromCtoA B.
ProduceC FuntilF Dis equal toC F, and then joinEandD.
In the trianglesE F C,E F D, what two sides were drawn equal?
What line is a side to each?
How great an angle isC F E?
What is a right angle?
Then how do the anglesC F EandE F Dcompare with each other?
If the two trianglesE F C,E F D, have the sideC Fof the one equal to the sideF Dof the other, the sideE Fcommon to both, and the included angleE F Cof the one equal to the included angleE F Dof the other, each to each, what do you infer?
Then what third side of the one have you found equal to a third side of the other?
C Eis what part of the broken lineC E D?
C Fis what part of the lineC D?
Which is shorter, the straight lineC D, or the broken lineC E D?
Then how does the half ofC DorC Fcompare with the half ofC E DorC E?
IfC Eis any line whatever other than a perpendicular, what may we now say of the perpendicular from the pointCto the straight lineA B?
We wish to prove that
A perpendicular is the shortest distance from a point to a straight line.
LetA Bbe a straight line, andC Apoint out of it; then will the perpendicularC Ebe the shortest line that can be drawn from the point to the line.
For draw any other line fromCtoA B, asC F.
ProduceC EuntilE DequalsC E, and joinF D.
The two trianglesF E C,F E D, have the sideC Eof the one equal to the sideE Dof the other, the sideF Ecommon, and the included angleF E Cof the one equal to the included angleF E Dof the other, they are therefore equal, and the sideC Fequals the sideF D.
But the straight lineC Dis the shortest distance between the two pointsC D; therefore it is shorter than the broken lineC F D.
ThenC E, the half ofC D, is shorter thanC F, the halfC F D.
And, asC Fis any line other than a perpendicular, the perpendicularC Eis the shortest line that can be drawn fromCtoA B.
We wish to prove that
A tangent to a circumference is perpendicular to a radius at the point of contact.
Let the straight lineA Bbe tangent at the pointDto the circumference of the circle whose centre isC.
Join the centreCwith the point of contactD, the tangent will be perpendicular to the radiusC D.
For draw any other line from the centre to the tangent, asC F.
As the pointDis the only one in which the tangent touches the circumference, any other point, asF, must be without the circumference.
Then the lineC F, reachingbeyondthe circumference, must be longer than the radiusC D, which would reach only to it; thereforeC Dis shorter than any other line which can be drawn from the pointCto the straight lineA B; therefore it is perpendicular to it.
PROPOSITION XVIII. THEOREM.
We wish to prove that
In any isosceles triangle, the angles opposite the equal sides are equal.
Let the triangleA B Cbe isosceles, having the sideA Bequal to the sideA C; then will the angleB, opposite the sideA C, be equal to the angleC, opposite the equal sideA B.
For draw the lineA Dso as to divide the angleAinto two equal parts, and let it be long enough to divide the sideB Cat some point asD.
Now the two trianglesA D B,A D C, have the sideA Bof the one equal to the sideA Cof the other, the sideA Dcommon to both, and the included angleB A Dof the one equal to the included angleC A Dof the other; therefore the two triangles are equal in all respects, and the angleB, opposite the sideA C, is equal to the angleC, opposite the sideA B.
We wish to prove that,
If two triangles have the three sides of the one equal to the three sides of the other, each to each, they are equal in all their parts.
Let the two trianglesA B C,A D C, have the sideA Bof the one equal to the sideA Dof the other; the sideB Cof the one equal to the sideD Cof the other, and the third side likewise equal; then will the two triangles be equal in all their parts.
For place the two triangles together by their longest side, and join the opposite verticesBandDby a straight line.
Because the sideA Bis equal to the sideA D, the triangleB A Dis isosceles, and the anglesA B D,A D B, opposite the equal sides are equal.
Because the sideB Cis equal to the sideD C, the triangleB C Dis isosceles, and the anglesC B D,C D B, opposite the equal sides are equal.
If to the angleA B Dwe add the angleD B C, we shall have the angleA B C.
And if to the equal ofA B D, that is,A B D, we add the equal ofD B C, that is,B D C, we shall have the angleA D C.
Therefore the angleA B Cis equal to the angleA D C.
Then the two trianglesA B C,A D C, have two sides, and the included angle of the one equal to two sides and the included angle of the other, each to each, and are equal in all their parts; that is, the three angles of the one are equal to the three angles of the other, and their areas are equal.
We wish to prove that
An angle at the circumference is measured by half the arc on which it stands.
LetB A Dbe an angle whose vertex is in the circumference of the circle whose centre isC; then will it be measured by half the arcB D.
For through the centre draw the diameterA E, and join the pointsCandB.
The exterior angleE C Bis equal to the sum of the anglesBandB A C.
Because the sidesC A,C B, are radii of the circle, they are equal, the triangle is isosceles, the anglesBandB A Copposite the equal sides are equal, and the angleB A Cis half of both.
Then, because the angleB A Cis half ofBandB A C, it must be half of their equalE C B.
ButE C B, being at the centre, is measured byB E; then half of it, orB A C, must be measured by halfB E.
In like manner, it may be proved that the angleC A Dis measured by halfE D.
Then, becauseB A Cis measured by halfB E, andC A Dby halfE D, the whole angleB A Dmust be measured by half the whole arcB D.
Suppose the angle were wholly on one side of the centre, asF A B.
Draw the diameterA Eand the radiusB Cas before.
Prove that the angleB A Eis measured by half the arcB E.
Draw another radius fromCtoF, and prove thatF A Eis measured by half the arcF E.
Then, because the angleF A Eis measured by half the arcF E, and the angleB A Eis measured by half the arcB E,
The difference of the angles, orF A B, must be measured by half the difference of the arcs, or half ofF B.
We wish to prove that
Parallel chords intercept equal arcs of the circumference.
Let the chordsA B,C D, be parallel; then will the intercepted arcsA CandB Dbe equal.
For draw the straight lineB C.
Because the linesA BandC Dare parallel, the interior alternate anglesA B C,B C D, are equal.
But the angleA B Cis measured by half the arcA C;
And the angleB C Dis measured by half the arcB D:
Then, because the angles are equal, the half arcs which measure them must be equal, and the whole arcs themselves must be equal.
We wish to prove that
The angle formed by a tangent and a chord meeting at the point of contact is measured by half the intercepted arc.
Let the tangentC A Band the chordA Dmeet at the point of contactA; then will the angleB A Dbe measured by half the intercepted arcA D.
For draw the diameterA E F.
BecauseA Bis a tangent, andA Ea radius at the point of contact, the angleB A Fis a right angle, and is measured by the semicircleA D F.
Because the angleF A Dis at the circumference, it is measured by half the arcD F.
Then the difference between the anglesB A FandD A F, orB A D, must be measured by half the difference of the arcsA D FandD F, orA D;
That is, the angleB A Dis measured by half the arcA D.