THEOREM VII.

Similar triangles are to one another as the squares described upon their homologous sides; (or in the duplicate ratio of their homologous sides).

A

B

C B

Α

Let ABC, A'B'C', be similar triangles, they shall be to one another as the

or

squares on BC and B'C'.

For since ABC, A'B'C' are similar triangles,

AB: BC= A'B: B'C'.

Therefore AB. BC: BC= A'B'. B'C': B'C'2;

AB. BC: A'B'. B'C' = BC2: B'C'2.

But since B = B',

the triangle ABC : triangle A'B'C' = AB. BC : A'B'. B'C'.

Therefore the

triangle ABC: triangle A'B'C' = BC2 : B′ C′′2.

NOTE. Duplicate ratio is thus defined:

[IV. 6.

When three magnitudes are proportionals, the first is said to have to the third the duplicate ratio of that which it has to the second.

Thus, if A: B=B: C, A has to C the duplicate ratio of that which it has to B. It will be seen that this ratio is equal to that of A2 to B2.

THEOREM VIII.

Similar parallelograms are to one another as the squares upon their homologous sides; (or in the duplicate ratio of their homologous sides).

Similar polygons are to one another in the duplicate ratio of their homologous sides.

For they may always be divided into a like number of similar triangles, each triangle in the first bearing a constant ratio to the corresponding triangle in the second.

The proof we leave as an exercise for the learner.

The proposition that "Similar figures are to one another in the duplicate ratio of their homologous sides. is true of curvilinear figures as well as of rectilinear. Thus two circles are to one another as the squares on their diameters, or on their radii. The proof of this is too difficult to be given here.

Properties of a Triangle.

THEOREM X.

In the triangle ABC, if AD be drawn from A to D the middle point of BC, and BE, CF be similarly drawn from B and C to the middle points of the opposite sides; AD, BE, CF shall intersect in one point.

A

B

Let AD and BE intersect in G. Join DE.

That is, the point of intersection of AD and BE is so situated on AD, that its distance from A is twice as great as its distance from D.

Similarly the point of intersection of AD, CF is so situated on AD, that its distance from A is twice as great as its distance from D.

Therefore these points coincide.

This is a proposition of great importance in Mechanics. The point G is the centre of inertia of the triangle ABC.

THEOREM XI.

In the triangle ABC, if AD be drawn from A perpendicular to BC, and BE, CF be similarly drawn from B and C perpendicular to the opposite sides; AD, BE, CF shall intersect in one point.

Through A, B, and C draw GH, HK, KG parallel to BC, CA, AB respectively.

The figures AK, BG, CH are parallelograms.

Similarly B and C bisect HK, KG.

Therefore every point on AD is equidistant from G

and H,

[I. 19.

and every point on BE is equidistant from H and K.

[I. 19.

Therefore the point of intersection of AD, BE is equidistant from G and K.

Therefore it is a point on CF.

THEOREM XII.

[1. 21.

If ABC be a right-angled triangle having a right angle at A, and from A, AD be let fall perpendicular to BC, the triangles ABD, ACD shall be similar to the whole triangle and to one another.

For in the triangles ABD, ABC the right angle ADB is equal to the right angle BAC, and the angle at B is common; therefore the triangles are equiangular and similar.