1.2. In triangle $A B C$, the altitude $A H$ and the bisector $A L$ were drawn. It turned out that $B H: H L: L C=1: 1: 4$ and side $A B$ is equal to 1. Find the lengths of the other sides of triangle $A B C$.
1.2. Answer: $A C=2, B C=3 / \sqrt{2}$. By the property of the angle bisector, we have $A C: A B=L C: L B=2$, from which $A C=2 A B=2$. Next, let $B H=x$, then $H C=5 x$. By the Pythagorean theorem for triangles $A B H$ and $A H C$, we get $1^{2}-x^{2}=$ $A H^{2}=2^{2}-25 x^{2}, 24 x^{2}=3, x=1 / 2 \sqrt{2}, B C=3 / \sqrt{2}$.