11. ABCD is a square. E, F, G, H are the middle points of AB, BC, CD, DA, and EF, FG, GH, HE are joined. Prove that EFGH has all its sides equal.

12. Prove by superposition that the squares described on two equal straight lines are equal.

13. If two quadrilaterals have three consecutive sides and the two contained angles in the one respectively equal to three consecutive sides and the two contained angles in the other, the quadrilaterals shall be equal in every respect.

PROPOSITION 5. THEOREM.

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

A

B

F

E

D

In A ABC, let AB AC, and let AB, AC be produced to

D and E:

=

it is required to prove ▲ ABC = L ACB and L DBC = L ECB.

In BD take any point F,

and from AE cut off AG join BG, CF.

=

I. 3 Post. 1

=

Const.

(1) In As AFC, AGB,

.. FC = GB, L AFC = ▲ AGB, ▲ ACF = ▲ ABG.

I. 4

(2) Because the whole AF = whole AG,

Const. Нур. I. Ax. 3

Proved in (2)

Proved in (1)

▲ CGB; Proved in (1)

LBCF = L CBG, and ▲ FBC

and the part

ABG
CBG

=

=

L

= LGCB.

I. 4

whole ACF, Proved in (1)

part BCF; Proved in (3)

.. the remainder ▲ ABC = remainder ▲ ACB; I. Ax. 3

and these are the angles at the base.

But it was proved in (3) that ▲ FBC

=

L GCB;

and these are the angles on the other side of the base.

COR.-If a triangle have all its sides equal, it will also have all its angles equal; or, in other words, if a triangle be equilateral, it will be equiangular.

1. If two angles of a triangle be unequal, the sides opposite to them will also be unequal.

2. Two isosceles triangles ABC, DBC stand on the same base BC, and on opposite sides of it; prove ▲ ABD = ▲ ACD.

3. Two isosceles triangles ABC, DBC stand on the same base BC, and on the same side of it; prove ▲ ABD = = L ACD.

4. In the figure to the second deduction, if AD be joined, prove that it will bisect the angles at A and D.

5. ABC is an isosceles triangle having AB = AC. In AB, AC, two points D, E are taken equally distant from A; prove that the triangles ABE, ACD are equal in all respects, and also the triangles DBC, ECB.

6. Prove that the opposite angles of a rhombus are equal.

7. D and E are the middle points of the sides BC and CA of a triangle; DO and EO are perpendicular to BC and CA; show that the angles OAB and OBA are equal.

8. Prove the proposition by supposing the ▲ ABC, after leaving a trace or impression of itself, to be lifted up, turned over, and applied to the trace.

9. Prove the first part of the proposition by supposing the angle at the vertex to be bisected.

PROPOSITION 6. THEOREM.

If two angles of a triangle be equal, the sides opposite them shall also be equal.

.. area of ▲ DBC : = area of ▲ ACB;

I. 4

which is impossible, since ▲ DBC is a part of ▲ ACB. Hence AC is not unequal to AB;

COR.-If a triangle have all its angles equal, it will also have all its sides equal; or, in other words, if a triangle be equiangular, it will be equilateral.

1. If two sides of a triangle be unequal, the angles opposite to them will also be unequal.

2. If ABC be an isosceles triangle, and if the equal angles ABC, ACB be bisected by BD, CD, which meet at D; prove that DBC is also an isosceles triangle.

3. In the figure to I. 5, if BG, CF intersect at H, prove that HBC is an isosceles triangle.

4. Hence prove that FH = GH, and that AH bisects

A.

5. By means of what is proved in the last deduction, give a method of bisecting an angle.

6. Prove the proposition by supposing the ▲ ABC, after leaving a trace or impression of itself, to be lifted up, turned over, and applied to the trace.

Two triangles on the same base and on the same side of it cannot have their conterminous sides equal.

If it be possible, let the two As ABC, ABD on the same base AB, and on the same side of it, have AC and BC BD.

=

AD,

Three cases may occur:

(1) The vertex of each ▲ may be outside the other A.

▲ may be inside the other A.
▲ may be on a side of the other ▲.
and in the second case join

(2) The vertex of one
(3) The vertex of one
In the first case join CD;
CD and produce AC, AD to E and F.

Because AC AD, .. ▲ ECD = ▲ FDC.

=

But ECD is greater than ▲ BCD;

=

I. 5 I. Ax. 8

BCD.

.. FDC is greater than ▲ BCD.
Much more then is ▲ BDC greater than
But because BC BD, .. ▲ BDC = ▲ BCD ;
BDC is greater than and equal to ▲ BCD,
which is impossible.

that is,

=

I. 5

BD.

The third case needs no proof, because BC is not Hence two triangles on the same base and on the same side of it cannot have their conterminous sides equal.

1. On the same base and on the same side of it there can be only one equilateral triangle.

2. On the same base and on the same side of it there can be only one isosceles triangle having its sides equal to a given straight line.

3. Two circles cannot cut each other at more than one point either above or below the straight line joining their centres.

PROPOSITION 8. THEOREM.

If three sides of one triangle be respectively equal to three sides of another triangle, the two triangles shall be equal in every respect; that is,

(1) The three angles of the one triangle shall be respectively equal to the three angles of the other triangle,

(2) The areas of the two triangles shall be equal.