And because AB falls along DE,

and the angle BAC is equal to the angle EDF, Hyp. therefore AC must fall along DF.

And because AC is equal to DF,

Hyp.

therefore the point C must coincide with the point F.
Then B coinciding with E, and C with F,
the base BC must coincide with the base EF;

for if not, two straight lines would enclose a space; which is impossible.

Ax. 10. Thus the base BC coincides with the base EF, and is therefore equal to it.

Ax. 8. And the triangle ABC coincides with the triangle DEF, and is therefore equal to it in area.

Ax. 8. And the remaining angles of the one coincide with the remaining angles of the other, and are therefore equal to them, namely, the angle ABC to the angle DEF,

and the angle ACB to the angle DFE.

That is, the triangles are equal in all respects.

Q. E. D.

NOTE. It follows that two triangles which are equal in their several parts are equal also in area; but it should be observed that equality of area in two triangles does not necessarily imply equality in their several parts: that is to say, triangles may be equal in area, without being of the same shape.

Two triangles which are equal in all respects have identity of form and magnitude, and are therefore said to be identically equal, or congruent.

The following application of Proposition 4 anticipates the chief difficulty of Proposition 5.

A

In the equal sides AB, AC of an isosceles triangle
ABC, the points X and Y are taken, so that AX
is equal to AY; and BY and CX are joined.
Shew that BY is equal to CX.

In the two triangles XAC, YAB,

XA is equal to YA, and AC is equal to AB; Hyp. that is, the two sides XA, AC are equal to the two sides YA, AB, each to each;

and the angle at A, which is contained by these sides, is common to both triangles:

B

therefore the triangles are equal in all respects;

so that XC is equal to YB.

1.4.

Q. E.D.

PROPOSITION 5. THEOREM.

The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced, the angles on the other side of the base shall also be equal to one another.

Let ABC be an isosceles triangle, having the side AB equal to the side AC, and let the straight lines AB, AC be produced to D and E:

then shall the angle ABC be equal to the angle ACB, and the angle CBD to the angle BCE.

Construction. In BD take any point F;

and from AE the greater cut off AG equal to AF the less. i. 3.

Join FC, GB.

Proof. Then in the triangles FAC, GAB,

Because

FA is equal to GA, and AC is equal to AB,

Constr.

Нур.

also the contained angle at A is common to the

two triangles;

therefore the triangle FAC is equal to the triangle GAB in

all respects;

1. 4.

that is, the base FC is equal to the base GB, and the angle ACF is equal to the angle ABG, also the angle AFC is equal to the angle AGB. Again, because the whole AF is equal to the whole AG, of which the parts AB, AC are equal, therefore the remainder BF is equal to the remainder CG.

Hyp.

Because

and FC is equal to GB,

Proved.

Proved.

also the contained angle BFC is equal to the contained angle CGB,

Proved.

therefore the triangles BFC, CGB are equal in all respects; so that the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG.

I. 4.

Now it has been shewn that the whole angle ABG is equal to the whole angle ACF,

and that parts of these, namely the angles CBG, BCF, are also equal;

therefore the remaining angle ABC is equal to the remaining angle ACB;

and these are the angles at the base of the triangle ABC. Also it has been shewn that the angle FBC is equal to the angle GCB;

and these are the angles on the other side of the base. Q.E.D.

COROLLARY.

also equiangular.

Hence if a triangle is equilateral it is

EXERCISES.

1. AB is a given straight line and C a given point outside it: shew how to find any points in AB such that their distance from C shall be equal to a given length L. Can such points always be found?

2. If the vertex C and one extremity A of the base of an isosceles triangle are given, find the other extremity B, supposing it to lie on a given straight line PQ.

3. Describe a rhombus having given two opposite angular points A and C, and the length of each side.

4. AMNB is a straight line; on AB describe a triangle ABC such that the side AC shall be equal to AN and the side BC to MB.

5. In Prop. 2 the point A may be joined to either extremity of BC. Draw the figure and prove the proposition in the case when A is joined to C.

H. E.

2

The following proof is sometimes given as a substitute for the first part of Proposition 5:

PROPOSITION 5. ALTERNATIVE PROOF.

A A

Β'

Let ABC be an isosceles triangle, having AB equal to AC :
then shall the angle ABC be equal to the angle ACB.

Suppose the triangle ABC to be taken up, turned over and laid down again in the position A'B'C', where A'B', A'C', B'C' represent the new positions of AB, AC, BC.

Then A'B' is equal to A'C' ; and A'B' is AB in its new position, therefore AB is equal to A'C';

in the same way AC is equal to A'B' ;

and the included angle BAC is equal to the included angle C'A'B', for they are the same angle in different positions;

I. 4.

therefore the triangle ABC is equal to the triangle A'C'B' in all respects : so that the angle ABC is equal to the angle A'C'B'. But the angle A'C'B' is the angle ACB in its new position; therefore the angle ABC is equal to the angle ACB.

Q. E.D.

EXERCISES.

CHIEFLY ON PROPOSITIONS 4 AND 5.

1. Two circles have the same centre O; OAD and OBE are straight lines drawn to cut the smaller circle in A and B and the larger circle in D and E prove that

2.

(i) AD=BE.

(ii) DB=AE.

(iii) The angle DAB is equal to the angle EBA.

(iv)The angle ADB is equal to the angle OEA.

ABCD is a square, and L, M, and N are the middle points of AB, BC, and CD: prove that

(i) LM = MN.

(ii) AM DM.

(iii) AN=AM.

(iv) BN DM.

[Draw a separate figure in each case].

3. O is the centre of a circle and OA, OB are radii; OM divides the angle AOB into two equal parts and cuts the line AB in M: prove that AMBM.

4. ABC, DBC are two isosceles triangles described on the same base BC but on opposite sides of it: prove that the angle ABD is equal to the angle ACD.

5. ABC, DBC are two isosceles triangles described on the same base BC, but on opposite sides of it: prove that if AD be joined, each of the angles BAC, BDC will be divided into two equal parts.

6. PQR, SQR are two isosceles triangles described on the same base QR, and on the same side of it: shew that the angle PQS is equal to the angle PRS, and that the line PS divides the angle QPR into two equal parts.

7. If in the figure of Exercise 5 the line AD meets BC in E, prove that BE EC.

8. ABCD is a rhombus and AC is joined: prove that the angle DAB is equal to the angle DCB.

9. ABCD is a quadrilateral having the opposite sides BC, AD equal, and also the angle BCD equal to the angle ADC: prove that BD is equal to AC.

10. AB, AC are the equal sides of an isosceles triangle; L, M, N are the middle points of AB, BC, and CA respectively: prove that LM=MN.

Prove also that the angle ALM is equal to the angle ANM.

DEFINITION. Each of two Theorems is said to be the Converse of the other, when the hypothesis of each is the conclusion of the other.

It will be seen, on comparing the hypotheses and conclusions of Props. 5 and 6, that each proposition is the converse of the other.

NOTE. Proposition 6 furnishes the first instance of an indirect method of proof, frequently used by Euclid. It consists in shewing that an absurdity must result from supposing the theorem to be otherwise than true. This form of demonstration is known as the Reductio ad Absurdum, and is most commonly employed in establishing the converse of some foregoing theorem.

It must not be supposed that the converse of a true theorem is itself necessarily true: for instance, it will be seen from Prop. 8, Cor. that if two triangles have their sides equal, each to each, then their angles will also be equal, each to each; but it may easily be shewn by means of a figure that the converse of this theorem is not necessarily true.