In $a$ triangle ABC, the median AD divides ∠BAC in the ratio 1 : 2. Extend AD to E such that EB is perpendicular AB. Given that BE = 3, BA = 4, find the integer nearest to BC2.

Step 1: Here, D is mid-point of BC, hence BD : CD = 1 : 1

Step 2: Let ∠BAD = , then ∠CAD = 2

Step 3: 3 24

Step 4: => tan   , and tan 2 

Step 5: 4 7

Step 6: Now, using cot $m$ – $n$ theorem in ABC

Step 7: 2cot = cot – cot2

Step 8: => cot  

Step 9: Now, using sine rule in ABD, we get

Step 10: BD sin 

Step 11: 4 sin(  –  )

Step 12: 4 \times 3 252  482

Step 13: => BD 

Step 14: 5 \times 48

Step 15: 25 2  48 2

Step 16: So, 4BD 2  BC 2   29.29

Step 17: Nearest integer is 29.