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.


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 ;-)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.