Bernhard Marks.

Marks' first lessons in geometry, objectively presented online

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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.

[Illustration]


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.

[Illustration]


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.

[Illustration]


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?

[Illustration]


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_.

[Illustration]


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_?

[Illustration]


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_.

[Illustration]


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_.

[Illustration]


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?

[Illustration]


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.

[Illustration]


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.

[Illustration]

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.

[Illustration]


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.

[Illustration]


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.

[Illustration]


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.

[Illustration]


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.

[Illustration]


PROPOSITION XXIII. THEOREM.


DEMONSTRATION.

We wish to prove that

_A tangent and chord parallel to it intercept equal arcs of the
circumference._

Let A B be tangent to the circumference at the point D, and let C F be
a chord parallel to the tangent; then will the intercepted arcs C D
and D F be equal.

For from the point of contact D, draw the straight line D C.

Because the tangent and chord are parallel, the interior alternate
angles A D C and D C F are equal.

But the angle A D C, being formed by the tangent D A and the chord D
C, is measured by half the intercepted arc D C;

And the angle D C F, being at the circumference, is measured by half
the arc on which it stands, D F:

Then, because the angles are equal, the half arcs which measure them
are equal, and the arcs themselves are equal.

[Illustration]


PROPOSITION XXIV. THEOREM.


DEMONSTRATION.

We wish to prove that

_The angle formed by the intersection of two chords in a circle is
measured by half the sum of the intercepted arcs._

Let the chords A B and C D intersect each other in the point E; then
will the angle B E D or A E C be measured by half the sum of the
arcs A C, B D.

For from the point C draw C F parallel to A B.

Because the chords A B and C F are parallel, the arcs A C, B F, are
equal.

Add each of these equals to B D, and we have B D plus A C equal to B D
plus B F; that is, the sum of the arcs B D, A C, is equal to the arc
F D.

Because the chords A B, C F, are parallel, the opposite exterior and
interior angles D E B, D C F, are equal.

But D C F is an angle at the circumference, and is therefore measured
by half the arc F D.

Then the equal angle D E B must be measured by half of the arc F D, or
its equal B D, plus A C.

[Illustration]


PROPOSITION XXV. THEOREM.


DEMONSTRATION.

We wish to prove that

_The angle formed by two secants meeting without a circle is measured
by half the difference of the intercepted arcs._

Let the secants A B, A C, intersect the circumference in the points D
and E; then will the angle B A C be measured by half the difference
between the arcs B C and D E.

For from the point D draw the chord D F parallel to E C.

Because A C and D F are parallel, the opposite exterior and interior
angles B D F and B A C are equal.

Because the chords D F, E C, are parallel, the arcs D E and F C are
equal.

If from the arc B C we take the arc D E, or its equal F C, we shall
have left the arc B F;

But the angle B D F, being at the circumference, is measured by half
the arc B F:

Then the equal of B D F, or B A C, must be measured by half the arc B
F, or half the difference between the intercepted arcs B C and D E.


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Online LibraryBernhard MarksMarks' first lessons in geometry, objectively presented → online text (page 5 of 6)