Theorem 6.9: Two triangles are similar if and only if the ratios the lengths of their corresponding sides are all the same.

Proof: This proof is in two parts. The first part is to show that if the angles are the same size that the ratios of the lengths of the sides are all the same. The second part is to show that if the ratios of the lengths of the sides are all the same then all of the corresponding angles are the same size.

We first show that if the angles are the same size that the ratios of the lengths of the sides are all the same.

In the illustration

• m/ A = m/ A'
• m/ B = m/ B'
• m/ C= m/ C'

Let   B"   be the point on the ray from   A   through   B   which is the same distance from   A   as   B'   is from   A',   and let   C"   be the point on the ray from   A   through   C   which is the same distance from   A   as   C'   is from   A'.

Triangle A"B"C"   is then congruent to   triangle A'B'C'   by SAS, Theorem 6.8, so since   / A"B"C"   is congruent to   / B',   they have the same size by Theorem 6.6. Similarly,   / A"C"B"   has the same size as   / C'.

As a result,   m / A"B"C" = m/ B,   and   m/ A"C"B" = m/ C,   so by Theorem 6.4,   B"C"   is parallel to   BC.

But by Theorem 2.9,   B"C" || BC   if and only if there exists a real number   t   such that

B" = (1 - t)A + tB

and

C" = (1 - t)A + tC

Then by Theorem 2.2,

|A"B"| = t|AB|

and

|A"C"| = t|AC|

so

Since isometries preserve distances we have that

|A"B"| = |A'B'|

and

|A"C"| = |A'C'|

so

and the sides in the triangles are proportional.

To show that

we could simply reach the conclusion by relabeling, but we can derive it directly. Let

A = (x0, y0),

B = (x1, y1),

and

C = (x2, y2)

Then

B" = (1 - t)A + tB

= (1 - t)(x0, y0) + t(x1, y1)

= ((1 - t)x0 + tx1, (1 - t)y0) + ty1)

and

C" = (1 - t)A + tC

= (1 - t)(x0, y0) + t(x2, y2)

= ((1 - t)x0 + tx2, (1 - t)y0) + ty2)

so

|B"C"| =

But   t   is a ratio of two distances, so it is positive, and we get

= t |BC| = |B'C'|
so

For the converse, suppose that the ratios of the lengths of the sides are all the same. Let that common ratio be   t.   That is

Then

|A'B'| = t|AB|

|A'C'| = t|AC|

and

|B'C'| = t|BC|

If we let

B" = (1 - t)A + tB

and

C" = (1 - t)A + tC

then we know the following things. First,

|A"B"| = t|AB| = |A'B'|

and

|A"C"| = t|AC| = |A'C'|

by Theorem 2.2. Moreover   B"C"   will be parallel to BC, by Theorem 2.9, so

m/ B = m/ B"

and

m/ C = m/ C"

by Theorem 6.4, and of course

/ A = / A"

so they have the same size.

Since   triangle A'B'C'   will be congruent to   triangle A"B"C"   by Theorem 6.7,   triangle A'B'C'   will have the same size angles as   triangle ABC.

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