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You are given a triangle $ABC$ where $AC = BC$.

Points $K$ and $L$ lie on $AB$ ($K$ belongs to $AL$). $\measuredangle KCL=\frac{1}{2}\measuredangle ACB$.

Prove that you can build a triangle from segments $AK$, $KL$, $BL$.

I can intuitively see that $KL < \frac{1}{2}AB$ and $KL$ is the longest among those three, but I have no idea how to prove it.

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Reflect $A$ by line $CK$ and denote the reflected point by $X$. Then $X$ is also $B$ reflected by $CL$, you can check it by $|CX|=|CA|=|CB|$ and by angles at the point $C$.

Now, $KLX$ is the desired triangle ;-)

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