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If the segment A'B' is tangent to the incircle of triangle ABC, and that segment AB = segment CM; then, what is the ratio of the area of the triangle ABC to the area of the small triangle A'B’C? enter image description here

Angle CAB is not necessarily a right angle...

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$S_{\triangle ABC}=\frac{1}{2}l_{\triangle ABC}\times r$

$S_{\triangle A'B'C}=\frac{1}{2}(l_{\triangle A'B'C}-2A'B')\times r$

the ratio=$\frac{S_{\triangle ABC}}{S_{\triangle A'B'C}}=\frac{l_{\triangle ABC}}{l_{\triangle A'B'C}-2A'B'}$

$l_{\triangle ABC}=2CM+2AB=4CM$

$l_{\triangle A'B'C}=2CM$

So, the ratio=$\frac{4CM}{2CM-2A'B'}=\frac{2CM}{CM-A'B'}$a triangle with its innercircle

While if $A'B'//AB $, the question becomes more easy.

the ratio=$\frac{S_{\triangle ABC}}{S_{\triangle A'B'C}}=(\frac{l_{\triangle ABC}}{l_{\triangle A'B'C}})^2=(\frac{4CM}{2CM})^2=4$

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