Theorem 6.9: If triangles are similar then the ratios the lengths of their corresponding sides are all the same.

Proof: We show that if the angles are congruent that the ratios of the lengths of the corresponding sides are all the same.

In the illustration

• / A/ D
• / B/ E
• / C/ F

Since   DA,   we can move   D   to be on top of   A,  by the definition of congruent. If   E   is not on   AB,   (or it's extension) then rotate   ΔDEF   about the angle bisector of   / D,   and we will be able to assume that it is. Then rotate the figure about   A   until   BC   is vertical.

Since AB and AC intersect BC , they are not vertical, and will, thus, have finite real slopes with which we can do arithmetic.

Since / AEF/ B,   EF || BC,   by Theorem 6.2, so   |EF|   is also vertical.

Let us define coordinates for these points. Let

A = D = (xo, yo)
B = (x1, y1)
C = (x2, y2)
E = (x3, y3)
F = (x4, y4)

Since   BC   is vertical,   x1 = x2,   and, since   EF   is vertical,   x3 = x4.   Let

m1   be the slope of   AB

and

m2 be the slope of   AC

Then by Theorem 1.5,

so

and

But, since   x1 = x2,   and   x3 = x4   This gives us

As required. Since   A   and   D   were an arbitrary pair of congruent angles, the same proof could be repeated for the other pairs of corresponding sides.

For the converse, assume that the corresponding sides in two triangles are proportional. In our example,

we are now assuming that

We can also assume, after, perhaps, relabeling, that

α < 1.

(If   α = 1,   then the triangles are congruent by SSS, and that angles are the same size because they would then be corresponding parts of congruent triangles.)

Let   B′   be the point on   AB   such that

|AB′| = α|AB

Draw the line through   B′   parallel to  AB.

Let   C′   be where this line intersects BC.

Since   BC′ || BC   ΔABC ∼ ΔABC′   so by the proof of the first part of the theorem, the sides in   ΔABC′   are proportional to the sides in   ΔABC.   As a result,

|AB′|= α|AB| = |DE|,

|AC′| = α|AC| = |DF|,

|BC′| = α|BC| = |EF|.

As a result,   ΔABC′ ≅ ΔDEF,   by SSS, and so they have the same angles. Conclude that   ΔDEF ∼ ΔABC.

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