**a**)
It is given:

*CM* = *AB* = *m*;

triangles *ABC* and *A’B’C* are
similar; so, tangent *A’B* is parallel
to the side *AB*.

__Triangle __*ABC*

**b**) According to the tangent property:
the lengths of intersecting tangents from their intersecting
points to their points of contact with the enclosed
circle are always equal, we have:

*CM* = *CN* (fig 1), and...

**c**) *BM* + *AN* = *AP* + *BP* = *AB*

**d**) Thus, the perimeter of the triangle *ABC* is:

*CM* + *BM* + *CN* + *AN* + *AB* =

2*CM* + (*BM* + *AN*) + *AB* =

2(*CM* + *AB*) = 4*CM* = 4*m*

__Small
triangle __*A’B’C*

**e**) As above, the lengths of intersecting
tangents from their intersecting points to their
points of contact with the enclosed circle are always
equal, therefore:

*A’N* = *A’Q* e *B’Q* = *B’M* (fig.
2)

**f**) Thus the perimeter of the small
triangle *A’B’C* is:

(*CM* - *B’M*) + (*CN* - *A’N*)
+ (*A’Q* + *B’Q*)

that is:

2*CM* - *B’M* - *A’N* + *A’Q* + *B’Q* =
2*CM* = 2*m*

__In
conclusion__

**g**) According to Euclid, if two triangles
are similar, then the ratio of their areas is the
square of the ratio of any two corresponding sides.

Then, the ratio of the area of triangle *ABC* /
area triangle *A’B’C* is:

(4*m*/2*m*)^{2} = 4

__Another
solution__

And here is an algebrical solution submitted by Fu
Su:

**a**) Semiperimeter *p'* of
quadrilateral *ABB'A'*:

*p'* = a + b + c + d (see drawing below)

**b**) Area of quadrilateral *ABB'A'*:

*r*(a + b + c + d)

**c**)
Semiperimeter *p* of triangle *ABC*:

a + b + (c + e) = 2(a + b) [since c + e = a + b]

**d**) Area of triangle *ABC*:
2*r*(a + b)

**e**)
Area of triangle *A'B'C*:

triangle *ABC* - quadrilateral *ABB'A'* =
2*r*(a + b) - *r*(a + b + c + d) =

= *r*[(a + b) - (c + d)]

**f**)
Then, Area of triangle *A'B'C* / Area of
triangle *ABC*:

*r*[(a + b) - (c + d)] / 2*r*(a + b)
=

= [(a + b) - (c + d)] / 2(a + b) =

= 1/2 - 1/2[(c + d) / (a + b)]

**g**) Also Area triangle *A'B'C* /
Area triangle *ABC* =

= (*A'B'*/*AB*)^{2} = [(c +
d) / (a + b)]^{2}

**h**)
Let, (c + d) / (a + b) = x,

then x^{2} = 1/2 - x/2 (see paragraph 'f')

or 2x^{2} + x - 1 = 0

hence (2x - 1)(x + 1) = 0 and x = 1/2 or -1 (cannot
be)

Area triangle *A'B'C*/Area triangle *ABC* =
x^{2} = **1/4**