Diagram 46.
DEMONSTRATION.
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 line e f intersect the two straight lines a b, c d, in the points g and h, so that the angles Red, Blue, are equal to two right angles.
Then will the lines a 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 line h d from the line e f.
And the angle Green measures the direction of the line g b from the line e f.
Then the lines g b, h d, have the same direction, and are parallel.
TEST LESSON.
1. Prove the same without the colors.
2. Prove the same, using the angle f h d.
3. Prove the same, supposing the angles a g h, g h c, equal to two right angles, and using the angle a g e.
4. Prove the same, using the angle c h f.
See Note E, 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:—
DEMONSTRATION.
We wish to prove that
The sum of any two sides of a triangle is greater than the third side.
Let the figure a b c be a triangle, then will the sum of any two sides, as a c, c b, be greater than the third side a b.
For the straight line a b is the shortest distance between the two points a and b, and is therefore less than the broken line a c b.
PROPOSITION IX. PROBLEM.
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.
SOLUTION.
Let a b be the given line.
With the point a as a centre, and a b as a radius, draw the circumference of the circle, or a part of one.
With the point b as a centre, and the same radius a b, draw another circumference, or a part of one.
From the point c, in which the circumferences or arcs intersect, draw the straight lines a c and b c.
Now, because the lines a b and a c are radii of the same circle, they are equal.
And, because the lines a b and b c are radii of the same circle, they are also equal.
Then, because the two lines a c, b c, are separately equal to the line a b, they are equal to each other, and the triangle is equilateral.
PROPOSITION X. THEOREM.
DEVELOPMENT LESSON.
Let the figure a b c be a triangle.
Produce the side a c to d.
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 point c draw the line c e parallel to a b.
Because the straight line a d intersects the two parallels a b, c e, the angle a is equal to what other angle?
Because the straight line b c intersects the two parallels a b, c e, the angle b is equal to what other angle?
Then the angles a and b are equal to what two angles?
How does the angle b c d compare with the angles b c e, e c d?
Then, if the angles a and b, on the one hand, and the angle b c d, on the other, are separately equal to the angles b 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?
DEMONSTRATION.
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.
Let a b c be a triangle.
Produce the side a c to d; then will the new angle b c d be equal to the sum of the angles a and b.
For from the point c draw c e parallel to a b.
Then, because the straight line a d intersects the two parallels a b, c e, in the points a and c,
The opposite exterior and interior angles a and e c d are equal to each other.
And because the straight line b c intersects the same parallels in the points b and c,
The interior alternate angles b and b c e are equal.
Then the angles a and b of the triangle are equal to the angles b c e and e c d.
But the new angle b c d is equal to the angles b c e, e c d.
Then because the new angle b c d, and the angles a and b are separately equal to the angles b c e, e c d, they are equal to each other.
PROPOSITION XI. THEOREM.
DEVELOPMENT LESSON.
Let the figure a b c be a triangle.
Produce the side a c to d.
By the last theorem, the angle b c d is 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 angle b c d, what adjacent angles do we get?
Then the three angles of the triangle, a, b, and c, are equal to what two angles?
But the adjacent angles a c b and b c d are equal to what?
Then, because the three angles of the triangle, a, b, and c, and two right angles, are separately equal to the two adjacent angles c and b 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 figure a b c be a triangle; then will the sum of the angles a, b, and c, be equal to two right angles.
For, produce the side a c to d.
The new angle b c d is equal to the sum of the angles a and b.
If to the angles a and b we add the angle c, we shall have the three angles of the triangle.
If to the angle b c d we add the same angle c, we shall have the adjacent angles c and b c d.
Then the three angles of the triangle a, b, c, are equal to the adjacent angles c and b c d.
But the adjacent angles c and b c d are equal to two right angles.
Then, because the three angles of the triangle are equal to the adjacent angles c and b c d, they are equal to two right angles.
PROPOSITION XII. THEOREM.
DEVELOPMENT LESSON.
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.
DEMONSTRATION.
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.
PROPOSITION XIII. THEOREM.
DEVELOPMENT LESSON.
In these two triangles we have tried to make the side a b of the one equal to the side d e of the other; the side a c of the one equal to the side d f of the other; and the included angle b a c of the one equal to the included angle e d f of the other.
We now wish to find out if the third side b c of the one is equal to the third side e f of the other, and if the two remaining angles b and c of the one are equal to the two remaining angles e and f of the other.
Suppose we were to cut the triangle d e f out of the page, and place it upon the triangle a b c, so that the line d e should fall upon the line a b, and the point d upon the point a.
As the line d e is equal to the line a b, upon what point will the point e fall?
If the angle e d f were less than the angle b a c, would the line d f fall within or without the triangle?
If the angle e d f were greater than the angle b a c, where would the line d f fall?
Since the angle a is equal to d, where, then, must the line d f fall?
As the line d f is equal to the line a c, upon what point will the point f fall?
Then, if the point e falls upon the point b, and the point f upon the point c, where will the line e f fall?
Now, because the three sides of the triangle d e f exactly fall upon the three sides of the triangle a b c, we say the two magnitudes coincide throughout their whole extent, and are therefore equal.
What three parts of the triangle a b c did we suppose to be equal to three corresponding parts of the triangle d e f before we placed one upon the other.
What line of the one do we find equal to a line in the other?
What two angles of the one do we find equal to two angles in the other?
What do you think of the areas of the triangles?
DEMONSTRATION.
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 triangles a b c and d e f have the side a b of the one equal to the side d e of the other; the side a c of the one equal to the side d f of the other; and the included angle b a c of the one equal to the included angle e d f of the other, each to each; then will the two triangles be equal in all their parts.
For, place the triangle d e f upon the triangle a b c, so that the line d e shall fall upon the line a b, with the point d upon the point a.
Because the line d e is equal to the line a b, the point e will fall upon the point b.
Because the angle e d f is equal to the angle b a c, the line d f will fall upon the line a c.
Because the line d f is equal to the line a c, the point f will fall upon the point c.
Then, because the point e is on the point b, and the point f on the point c, the line e f will coincide with the line b c, and the two triangles will be found equal in all their parts;
That is, the angle e is found to be equal to the angle b, the angle f to the angle c, the line e f to the line b c, and the area of the triangle a b c to the area of the triangle d e f.
PROPOSITION XIV. THEOREM.
DEVELOPMENT LESSON.
In these two triangles we have tried to make the angle b of the one equal to the angle e of the other; the angle c of the one equal to the angle f of the other; and the included side b c of the one equal to the included side e f of the other.
We now wish to find out if the remaining angle a of the one is equal to the remaining angle d of the other, and if the two remaining sides a b and a c of the one are equal to the two remaining sides d e and d f of the other.
Suppose we were to cut the triangle d e f out of the page and place it upon the triangle a b c, so that the line e f shall fall upon the line b c, with the point e upon the point b.
Because the line e f is equal to the line b c, upon what point will the point f fall?
Because the angle e is equal to the angle b, where will the line e d fall?
Because the angle f is equal to the angle c, where will the line d f fall?
Then, if the line d e falls upon the line a b and the line d f upon the line a c, where will the point d fall?
Now because the three sides of the triangle d e f exactly fall upon the three sides of the triangle a b c, we say the two magnitudes coincide throughout their whole extent, and are therefore equal.
Suppose the angle e were greater than the angle b, would the line e d fall within or without the triangle?
If it were less, where would the line fall?
Why does the line d e fall exactly upon the line a b?
DEMONSTRATION.
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 triangles a b c and d e f have the angle b of the one equal to the angle e of the other; the angle c of the one equal to the angle f of the other; and the included side b c of the one equal to the included side e f of the other, each to each; then will the two triangles be equal in all their parts.
For place the triangle d e f upon the triangle a b c, so that the line e f shall fall upon the line b c, with the point e upon the point b.
Because the line e f is equal to the line b c the point f will fall upon the point c.
Because the angle e is equal to the angle b, the line e d will fall upon the line b a, and the point d will be somewhere in the line b a.
Because the angle f is equal to the angle c, the line f d will fall upon the line c a, and the point d will be somewhere in the line c a.
Then, because the point d is in the two lines, b a and c a, it must be in their intersection, or upon the point a.
And, as the two triangles coincide throughout their whole extent, they are equal in all their parts.
That is, the angle a is found to be equal to the angle d; the side b a to the side e d; the side c a to the side f d; and the area of the triangle a b c to the area of the triangle d e f.
PROPOSITION XV. THEOREM.
DEMONSTRATION.
We wish to prove that
The opposite sides of any parallelogram are equal.
Let the figure a b c d be a parallelogram; then will the sides a b and c d be equal to each other; likewise the sides a d and b c.
For, draw the diagonal b d.
Because the figure is a parallelogram, the sides a b and d c are parallel, and the interior alternate angles n and o are equal.
Because the figure is a parallelogram, the interior alternate angles r and m are equal.
Then the two triangles a 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 side a b opposite the angle m is equal to the side c d opposite the equal angle r;
And the side a d opposite the angle n is equal to the side b c opposite the equal angle o.
TEST.
Prove the same by drawing a diagonal from a to c.
PROPOSITION XVI. THEOREM.
DEVELOPMENT LESSON.
Suppose A B to be a straight line, and C any point out of it.
From the point C draw a perpendicular C F to A 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 from C to A B as C E.
Now, as C E is any line whatever other than a perpendicular, if we find that the perpendicular C F is shorter than it we must conclude that it is the shortest line that can be drawn from C to A B.
Produce C F until F D is equal to C F, and then join E and D.
In the triangles E F C, E F D, what two sides were drawn equal?
What line is a side to each?
How great an angle is C F E?
What is a right angle?
Then how do the angles C F E and E F D compare with each other?
If the two triangles E F C, E F D, have the side C F of the one equal to the side F D of the other, the side E F common to both, and the included angle E F C of the one equal to the included angle E F D of 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 E is what part of the broken line C E D?
C F is what part of the line C D?
Which is shorter, the straight line C D, or the broken line C E D?
Then how does the half of C D or C F compare with the half of C E D or C E?
If C E is any line whatever other than a perpendicular, what may we now say of the perpendicular from the point C to the straight line A B?
DEMONSTRATION.
We wish to prove that
A perpendicular is the shortest distance from a point to a straight line.
Let A B be a straight line, and C A point out of it; then will the perpendicular C E be the shortest line that can be drawn from the point to the line.
For draw any other line from C to A B, as C F.
Produce C E until E D equals C E, and join F D.
The two triangles F E C, F E D, have the side C E of the one equal to the side E D of the other, the side F E common, and the included angle F E C of the one equal to the included angle F E D of the other, they are therefore equal, and the side C F equals the side F D.
But the straight line C D is the shortest distance between the two points C D; therefore it is shorter than the broken line C F D.
Then C E, the half of C D, is shorter than C F, the half C F D.
And, as C F is any line other than a perpendicular, the perpendicular C E is the shortest line that can be drawn from C to A B.
PROPOSITION XVII. THEOREM.
DEMONSTRATION.
We wish to prove that
A tangent to a circumference is perpendicular to a radius at the point of contact.
Let the straight line A B be tangent at the point D to the circumference of the circle whose centre is C.
Join the centre C with the point of contact D, the tangent will be perpendicular to the radius C D.
For draw any other line from the centre to the tangent, as C F.
As the point D is the only one in which the tangent touches the circumference, any other point, as F, must be without the circumference.
Then the line C F, reaching beyond the circumference, must be longer than the radius C D, which would reach only to it; therefore C D is shorter than any other line which can be drawn from the point C to the straight line A B; therefore it is perpendicular to it.
PROPOSITION XVIII. THEOREM.
DEMONSTRATION.
We wish to prove that
In any isosceles triangle, the angles opposite the equal sides are equal.
Let the triangle A B C be isosceles, having the side A B equal to the side A C; then will the angle B, opposite the side A C, be equal to the angle C, opposite the equal side A B.
For draw the line A D so as to divide the angle A into two equal parts, and let it be long enough to divide the side B C at some point as D.
Now the two triangles A D B, A D C, have the side A B of the one equal to the side A C of the other, the side A D common to both, and the included angle B A D of the one equal to the included angle C A D of the other; therefore the two triangles are equal in all respects, and the angle B, opposite the side A C, is equal to the angle C, opposite the side A B.
PROPOSITION XIX. THEOREM.
DEMONSTRATION.
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 triangles A B C, A D C, have the side A B of the one equal to the side A D of the other; the side B C of the one equal to the side D C of 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 vertices B and D by a straight line.
Because the side A B is equal to the side A D, the triangle B A D is isosceles, and the angles A B D, A D B, opposite the equal sides are equal.
Because the side B C is equal to the side D C, the triangle B C D is isosceles, and the angles C B D, C D B, opposite the equal sides are equal.
If to the angle A B D we add the angle D B C, we shall have the angle A B C.
And if to the equal of A B D, that is, A B D, we add the equal of D B C, that is, B D C, we shall have the angle A D C.
Therefore the angle A B C is equal to the angle A D C.
Then the two triangles A 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.
PROPOSITION XX. THEOREM.
DEMONSTRATION.
We wish to prove that
An angle at the circumference is measured by half the arc on which it stands.
Let B A D be an angle whose vertex is in the circumference of the circle whose centre is C; then will it be measured by half the arc B D.
For through the centre draw the diameter A E, and join the points C and B.
The exterior angle E C B is equal to the sum of the angles B and B A C.
Because the sides C A, C B, are radii of the circle, they are equal, the triangle is isosceles, the angles B and B A C opposite the equal sides are equal, and the angle B A C is half of both.
Then, because the angle B A C is half of B and B A C, it must be half of their equal E C B.
But E C B, being at the centre, is measured by B E; then half of it, or B A C, must be measured by half B E.
In like manner, it may be proved that the angle C A D is measured by half E D.
Then, because B A C is measured by half B E, and C A D by half E D, the whole angle B A D must be measured by half the whole arc B D.
SECOND CASE.
Suppose the angle were wholly on one side of the centre, as F A B.
Draw the diameter A E and the radius B C as before.
Prove that the angle B A E is measured by half the arc B E.
Draw another radius from C to F, and prove that F A E is measured by half the arc F E.
Then, because the angle F A E is measured by half the arc F E, and the angle B A E is measured by half the arc B E,
The difference of the angles, or F A B, must be measured by half the difference of the arcs, or half of F B.
PROPOSITION XXI. THEOREM.
DEMONSTRATION.
We wish to prove that
Parallel chords intercept equal arcs of the circumference.
Let the chords A B, C D, be parallel; then will the intercepted arcs A C and B D be equal.
For draw the straight line B C.
Because the lines A B and C D are parallel, the interior alternate angles A B C, B C D, are equal.
But the angle A B C is measured by half the arc A C;
And the angle B C D is measured by half the arc B D:
Then, because the angles are equal, the half arcs which measure them must be equal, and the whole arcs themselves must be equal.
PROPOSITION XXII. THEOREM.
DEMONSTRATION.
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 tangent C A B and the chord A D meet at the point of contact A; then will the angle B A D be measured by half the intercepted arc A D.
For draw the diameter A E F.
Because A B is a tangent, and A E a radius at the point of contact, the angle B A F is a right angle, and is measured by the semicircle A D F.
Because the angle F A D is at the circumference, it is measured by half the arc D F.
Then the difference between the angles B A F and D A F, or B A D, must be measured by half the difference of the arcs A D F and D F, or A D;
That is, the angle B A D is measured by half the arc A D.