Let A be the given point, and BC the given straight line.

It is required to draw from the point A, a straight line equal to BC.

H

B

A

From the point A to C draw the straight line AC (post. 1); upon AC describe the equilateral triangle ACD (1. 1); from the centre C, at the distance CB, describe the circle BGH (post. 3), cutting CD in the point G:

and from the centre D, at the distance DG, describe the circle GKL, cutting DA in the point L.

Then the straight line AL shall be equal to BC.

Proof. Because the point C is the centre of the circle BGH, therefore CB is equal to CG (def. 15);

and because D is the centre of the circle GKL,

therefore DL is equal to DG,

and DA, DC are equal (I. 1);

therefore the remainder AL is equal to the remainder CG (ax. 3): but it has been shown that CB is equal to CG,

wherefore AL and CB are each of them equal to CG;

and things which are equal to the same thing are equal to one another;

therefore the straight line AL is equal to BC (ax. 1). Wherefore from the given point A, a straight line AL has

been drawn equal to the given straight line BC. Q.E.F.

4. If a triangle ABC be turned over about its side AB, show that the line joining the two positions of C is perpendicular to AB.

Let the triangle ABC be turned over about its side AB, as ABD, and let CD be joined.

Then CD shall be perpendicular to AB.

Proof. Because the side CA is equal to the side DA (hyp.), and the side AE is common to the two triangles CAE, DAE, the two sides CA, AE are equal to the two sides DA, AE, each to each;

and the angle CAE is equal to the angle DAE (hyp.); therefore the base CE is equal to the base DE (1. 4), and the triangle CAE is equal to the triangle DAE,

also the remaining angles of the one are equal to the remaining angles of the other, each to each, to which the equal sides. are opposite;

therefore the angle AEC is equal to the angle AED;

and these two angles are adjacent angles;

therefore AE is perpendicular to CD (def. 10).

Wherefore CD is perpendicular to AB.

5. What is a plane triangle?

Q.E.D.

A plane triangle is the figure contained by three straight lines.

Show that every equilateral triangle is equiangular.

Let ABC be an equilateral triangle.

Then the triangle ABC shall be equiangular.

B

C

Proof. Because AB is equal to AC (hyp.),

therefore the angle ABC is equal to the angle ACB (1. 5); and because BA is equal to BC,

therefore the angle BAC is equal to the angle BCA or ACB: but the angle ABC is also equal to the angle ACB;

therefore the angle ABC is equal to the angle BAC (ax. 1). Wherefore the equilateral triangle ABC is equiangular. Q.E.D.

6. What property of equiangular triangles can be derived from this proposition? Give the steps necessary to prove this property. It can be proved by this proposition that equiangular triangles are also equilateral.

First prove that two sides are equal, then prove that one of those sides and the third side are equal

Wherefore (by ax. 1) it can be proved that it is equilateral.

A line which bisects the vertical angle of a triangle also bisects the base. Prove that the triangle is isosceles.

Let the straight line AD bisect the vertical angle BAC, and also the base BC of the triangle ABC.

Then the triangle ABC shall be isosceles.

B

D

Produce AD to E (post. 2), making DE equal to DA (1. 3),

and join CE (post. 1).

Proof. In the triangles ADB, CDE,

the side BD is equal to the side CD (hyp.),

the side AD is equal to the side ED (con.),

and the angle BDA is equal to the angle CDE (L. 15), therefore the base AB is equal to the base EC (1. 4), and the angle BAD is equal to the angle CED:

but the angle BAD is equal to the angle CAD (hyp.) ; therefore the angle CED is equal to the angle CAD (ax. 1), and therefore the side CE is equal to the side CA (1. 6) : but CE has been proved equal to AB;

therefore AB is equal to AC (ax. 1).

Wherefore the triangle ABC is isosceles. Q.E.D.

What proposition is the converse of this?

Prop. 5 (part of it) is the converse of prop. 6.

Prop. 6 gives two angles equal to prove the sides subtending the equal angles are equal; while prop. 5 gives two sides of a triangle equal to prove that the angles subtended by the equal sides are equal.

Show that every equiangular triangle is equilateral.
Let ABC be an equiangular triangle.

Then the triangle ABC shall be equilateral.

Proof.-Because the angle ABC is equal to the angle ACB (hyp.),

therefore the side AB is equal to the side AC (1. 6);

and because the angle BAC is equal to the angle BCA, therefore the side BA is equal to the side BC:

but it has been proved that the side AB is equal to the side AC;

therefore the side AC is equal to the side BC (ax. 1).

Wherefore the equiangular triangle ABC is equilateral. Q.E.D.

8. Is the equality of the triangles in all respects proved in this proposition? If not, how would you prove it?

The equality of the triangles in all respects is not proved; it can be proved by I. 4.

If two circles cut each other, the line joining their points of intersection is perpendicular to the line joining their centres.

Let the two circles CDE, CDF intersect at the points C, D, and let CD be joined, also A, B, the centres of the circles. Then CD shall be perpendicular to AB.

Let G be the point of intersection of AB, CD, and draw AC, AD, BC, BD (post. 1).

Proof. In the triangles CAB, DAB,

the side CA is equal to the side DA (def. 15),

the side AB is common to the two triangles,

and the base BC is equal to the base BD (def. 15), therefore the angle CAB is equal to the angle DAB (1. 8). Again, in the triangles CAG, DAG, the two sides CA, AG are equal to the two sides DA, AG, each to each,

and the angle CAG has been proved equal to the angle DAG,