B2. In triangle $A B C$, the bisector of angle $\angle B A C$ intersects side $B C$ at point $D$. Triangle $A D C$ is isosceles with vertex $D$, and it is also given that $|C D|=36$ and $|B D|=64$. Calculate the lengths of the sides of triangle $A B C$.
B2. Obviously, $|BC|=100$. Let's calculate the lengths of the other sides. Since $\angle BAD = \angle DAC = \angle ACB$, triangles $ABD$ and $CBA$ are similar. Therefore, $\frac{|AC|}{|AD|} = \frac{|AB|}{|BD|} = \frac{|BC|}{|AB|}$. From the second equality, we get $|AB| = \sqrt{|BD| \cdot |BC|} = \sqrt{64 \cdot 100} = 80$. From the first equality, we then get $|AC| = \frac{|AD| \cdot |AB|}{|BD|} = \frac{36 \cdot 80}{64} = 45$.

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Calculation of $|BC|=100$ ..... 1 point
Triangles $ABD$ and $CBA$ are similar ..... 2 points
Writing the ratio $\frac{|AB|}{|BD|} = \frac{|BC|}{|AB|}$ ..... 1 point
Calculation of $|AB|=80$ ..... 1 point
Writing the ratio $\frac{|AB|}{|BD|} = \frac{|AC|}{|AD|}$ and calculation of $|AC|=45$ ..... 1 point