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**a**) It is given:

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

triangles *ABC* and *Aâ€™Bâ€™C* are similar; so, tangent *Aâ€™B* is parallel to the side *AB*.[/FONT]

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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*

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Small triangle *Aâ€™Bâ€™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:

*Aâ€™N* = *Aâ€™Q* e *Bâ€™Q* = *Bâ€™M* (fig. 2)

**f**) Thus the perimeter of the small triangle *Aâ€™Bâ€™C* is:

(*CM* - *Bâ€™M*) + (*CN* - *Aâ€™N*) + (*Aâ€™Q* + *Bâ€™Q*)

that is:

2*CM* - *Bâ€™M* - *Aâ€™N* + *Aâ€™Q* + *Bâ€™Q* = 2*CM* = 2*m*[/FONT]

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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 *Aâ€™Bâ€™C* is:

(4*m*/2*m*)2 = 4[/FONT][FONT=Verdana, Arial, Helvetica, sans-serif]

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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)[/FONT]

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**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)[/FONT]

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**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)][/FONT]

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**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[/FONT]

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

then x2 = 1/2 - x/2 (see paragraph 'f')

or 2x2 + 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* = x2 = **1/4**

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