4. In triangle $A B C$, the bisector $B E$ and the median $A D$ are perpendicular and have the same length, equal to 60. Find the sides of triangle $A B C$.

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Solution. Let $P$ be the intersection point of segments $B E$ and $A D$. Triangle $A B D$ is isosceles because its bisector $B P$ is also an altitude. Therefore, $A P = P D = 30 ; B C = 2 B D = 2 A B$.

By the property of the bisector of a triangle, $\frac{C E}{A E} = \frac{B C}{A B} \Rightarrow \frac{C E}{A E} = 2 \Rightarrow A C = 3 A E$.

Doubling the median $A D$ of triangle $A B C (A D = D K)$. Note that $A B K C$ is a parallelogram and $B K = A C = 3 A E$.

From the similarity of triangles $A P E$ and $K P B$, it follows that $\frac{P E}{B P} = \frac{A E}{B K} = \frac{1}{3}$. Therefore, $P E = 15$ and $P B = 45$. Hence, $A B = \sqrt{A P^{2} + B P^{2}} = 15 \sqrt{13}$, then $B C = 2 A B = 30 \sqrt{13}$.

Furthermore, $A E = \sqrt{A P^{2} + E P^{2}} = 15 \sqrt{5}$, from which $A C = 3 A E = 45 \sqrt{5}$. Answer: $15 \sqrt{13}, 30 \sqrt{13}$, $45 \sqrt{5}$.