PROP. VII. THEOREM.

Upon the same base, and on the same side of it, there cannot be two triangles that have their sides terminated in one extremity of the base equal to each other, and likewise those which are terminated in the other extremity of the base equal to each other.

Suppose it possible that on the same base, AB, and on the same side of it-e.g. above it-there can be two triangles, ACB and ADB, having the sides AC and AD terminated in one extremity of the base, A,= each other, and having also the sides BC and BD terminated in the other extremity of the base, B,= each other;

Then this supposition will present itself in three cases. CASE I. Where the vertex of each triangle falls without the other, as in the following figure.

D

Again, because the

CONSTRUCTION.-Join the ver

tices CD.

PROOF. Because in the triangle ACD the side AC = the side AD (hyp.), therefore the angle ACD = the angle ADC (I. 5).

But the angle ACD is greater than the angle BCD(ax. 9); therefore the angle ADC is also greater than the angle BCD.

angle BDC is greater than the angle ADC (ax. 9), therefore the angle BDC is much greater than the angle BCD.

Next, because in the triangle BDC the side BD = the side BC (hyp.), therefore the angle BDC the angle BCD (I. 5). But we have just proved that the angle BDC is "much greater" than the angle BCD.

Therefore, on the above supposition, the angle BDC is both equal to, and greater than, the angle BCD, which is absurd.

Therefore the above supposition is false as referred to CASE I., where the vertex of one triangle falls without the other.

We now pass on to consider this same supposition under

CASE II-Where the vertex of one triangle, D, falls within the other, as in the following figure.

CONSTRUCTION.-1. Join the vertices CD.

2. Produce AC and AD to E and F respectively.

PROOF.-Because in the triangle ACD, the side AC-the side AD (hyp.) and these sides are produced to E and F respectively, therefore the angle ECD = the angle FDC (I. 5).

But the angle ECD is greater than the angle BCD (ax. 9); therefore the angle FDC is also greater than the angle BCD.

Again, because the angle BDC

A

E

F

is greater than the angle FDC (ax. 9), therefore the angle BDC is much greater than the angle BCD.

Next, because in the triangle BDC, the side BD = the side BC (hyp.), therefore the angle BDC= the angle BCD (I. 5). But we have just proved that the angle BDC is "much greater" than the angle BCD.

Therefore, on the above supposition, the angle BDC is both equal to, and greater than, the angle BCD, which is absurd.

Therefore the above supposition is false as referred to CASE II.,where the vertex of one triangle falls within the other.

CASE III-Where the vertex of one triangle falls on a side of the other, as in the following figure.

In this case it is evident that if one pair of the sides which are terminated in one extremity of the base, e g. AC and AD, terminated in A, equal each other, then the other pair, BC and

BD, terminated in B, A cannot be equal to each

D

B

other, as the Proposition requires, and therefore this case is

to be dismissed.

Wherefore,

Upon the same base, &c.

Q. E. D.

PROP. VIII. THEOREM.

If two triangles have the three sides of the one equal to the three sides of the other, each to each, then the angle which is contained by any two sides of the one triangle shall be equal to the angle contained by the two sides, equal to them, of the other.

In the triangles ABC and DEF, let the sides AB, BC, and CA in the former the sides DE, EF, and FD in the latter, each to each.

PROOF.-1. If the triangle ABC be placed upon the triangle DEF, so that the point B is on the point E, and the side BC upon the side EF, then, because BC = EF (hyp.), therefore the point C will coincide with the point F, and because the point B coincides with the point E, and the point C with the point F, therefore the side BC coincides with the side EF.

Next, because the side BC coincides with the side EF therefore the sides BA and AC shall coincide with the sides ED and DF respectively.

For, if BC coincides with EF, and then BA and AC do not coincide with ED and DF, suppose that BA and AC have another direction, as EG and GF, then, if this be true, we shall have upon the same base EF, and upon the same side of it, two triangles in a manner which is impossible (I. 7).

Therefore, if BC coincides with EF, then BA and AC must coincide with ED and DF, and the angle BAC will coincide with, and equal, the angle EDF (ax. 8).

Therefore, it is proved, as required, that

N.B.-The equality of the two triangles, in every respect, follows from this Proposition, as it does from Prop. IV.

Exercises.

1. In the fig. Prop. I. if E be the point where the circles intersect below AB, and AE and BE be joined; prove that the angle ACB: = the angle AEB.

2. If BC be the base of an isosceles triangle with vertical angle at A; prove that if a line AD bisects the base BC, it bisects also the vertical angle BAC.

To bisect a given rectilineal angle, i.e. to divide it into two equal parts.

Let BAC be the given rectilineal angle:

It is required to bisect it.

CONSTRUCTION.-1. In AB take any point, D.

2. From AC cut off AE =

AD (I. 3).

3. On DE construct an equilateral triangle DEF (I. 1). 4. Join AF.

Then it is to be proved that

The rectilineal angle BAC is bisected by the line AF.

PROOF.-Because in the two triangles DAF and EAF, we have the three sides DA, AF, and FD in the former =the three sides EA, AF, and FE, in the latter, each to each (cons.), therefore the angle DAF the angle EAF (I. 8), i.e. the angle BAF the angle CAF (note 2 def. 15).

=

=

Therefore, it is proved, as required, that

The rectilineal angle BAC is bisected by the line AF.

Q. E. F.

Exercise.

In the fig. Prop. I. if E be the point where the circles intersect below AB, and AE and BE be joined; prove that AB bisects the angle CBE.