A segment tangent to the incircle.

Question

$D$ and $E$ are points on sides $AB$ and $AC$ of a triangle $ABC$ such that $DE$ is parallel to $BC$ and tangent to the incircle $I$. Prove that $$DE\leq\frac{AB+BC+AC}{8}$$

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This looks like $AM-GM$, but I don't have a cube on $DE$. I tried using the triangle inequality like this:

I rotated $ABC$ by $180^\circ$ around the incentre, then applied the triangle inequality to get $$AD+AE+B'E+B'K+CK+CL\gt DE+EK+KL$$

Now, if this relation is true it should apply for all of $DE, EK,KL$.

I'm not sure where to go from here.

Please help. Hints will be appreciated.

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I just saw this, and it seems to be related, but different.

Answer 1

Ahh, it is much easier. Since $\Delta ABC\sim \Delta ADE$ we have $${\text{Perimeter}_{ADE}\over \text{Perimeter}_{ABC}} = {DE \over BC}$$ Let $a=y+z$, then $${x\over a+x} = {DE\over a}\implies DE = {ax\over a+x}$$ and we have to check: $${ax\over a+x} \leq {2a+2x\over 8}$$ which is easy to see is true.

Answer 2

Using tangent segments we get this picture.

Since $\Delta ABC\sim \Delta ADE$ we have $${d+e\over z+y}={x-d\over x+z}={x-e\over x+y} \;\;\;\;\;\;\;\;(*)$$ Taking first two fractions we get $$d(x+y+2z)+e(x+z)= x(z+y)$$ and taking 1.st and 3.rd fractions we get $$e(x+y+2z)+d(x+y)= x(z+y)$$ so after substracting these 2 equations we get $dz=ye$. Plugging $e = dz/y$ in $(*)$ we get $$ d= {xy\over x+y+z}$$ In the same manner we get $$e= {xz\over x+y+z}$$ Now we have to check this: $$ {8x(y+z)\over x+y+z}\leq 2x+2y+2z$$ which is equivalent to $4ax\leq (a+x)^2$ is true ($a=y+z$).