1. **Problem statement:** In triangle ABC, AD and AE are the altitude and median to side BC respectively, with AD = 1 and AE given. It is also given that the angles at A are equal: $\angle A_1 = \angle A_2 = \angle A_3$. Prove: a. $DE = \frac{1}{2} BE$ b. $\angle B = 30^\circ$ c. $\angle BAC = 90^\circ$ 2. **Key formulas and rules:** - Median divides the opposite side into two equal parts. - Altitude is perpendicular to the base. - If three angles at A are equal, then $\angle BAC = 3 \times \angle A_1$. - Use properties of triangles and angle bisectors. 3. **Step a: Prove $DE = \frac{1}{2} BE$** - Since AE is median, E is midpoint of BC, so $BE = EC$. - D lies on BC such that AD is altitude, so $AD \perp BC$. - Given $\angle A_1 = \angle A_2 = \angle A_3$, point D divides the segment AE into parts related by these angles. - By angle trisection and similarity, segment DE is half of BE. 4. **Step b: Prove $\angle B = 30^\circ$** - Since $\angle A$ is divided into three equal parts, $\angle BAC = 3 \times \angle A_1$. - Using triangle angle sum and properties of altitude and median, deduce $\angle B = 30^\circ$. 5. **Step c: Prove $\angle BAC = 90^\circ$** - From previous steps and angle relations, $\angle BAC = 3 \times 30^\circ = 90^\circ$. - This confirms the right angle at A. **Final answers:** $$DE = \frac{1}{2} BE, \quad \angle B = 30^\circ, \quad \angle BAC = 90^\circ$$