# We have a <mi mathvariant="normal">&#x25B3;<!-- △ --> A B C and a <mi mathvariant

We have a $\mathrm{△}ABC$ and a $\mathrm{△}{A}_{1}{B}_{1}{C}_{1}$. The segments $CL$ and ${C}_{1}{L}_{1}$ are angle bisectors. If $\mathrm{△}ALC\sim \mathrm{△}{A}_{1}{L}_{1}{C}_{1}$, I should show that $\mathrm{△}ABC\sim \mathrm{△}{A}_{1}{B}_{1}{C}_{1}$.

From the similarity, we have $\frac{AL}{{A}_{1}{L}_{1}}=\frac{CL}{{C}_{1}{L}_{1}}=\frac{AC}{{A}_{1}{C}_{1}}$. The only way I see from here is to show that $\mathrm{△}LBC\sim \mathrm{△}{L}_{1}{B}_{1}{C}_{1}$. Is this necessary for the solution?
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Kosyging1j7u
If $\mathrm{△}ALC\sim \mathrm{△}{A}_{1}{L}_{1}{C}_{1}$ then $\mathrm{\angle }A=\mathrm{\angle }{A}_{1}$. Also, $\mathrm{\angle }ACL=\mathrm{\angle }{A}_{1}{C}_{1}{L}_{1}$. But $\mathrm{\angle }C=2\mathrm{\angle }ACL$ and $\mathrm{\angle }{C}_{1}=2\mathrm{\angle }{A}_{1}{C}_{1}{L}_{1}$, so $\mathrm{\angle }C=\mathrm{\angle }{C}_{1}$, etc.