Discussion

Explanation:

​​​Here D is midpoint of AC, E is the midpoint of AD and F is the midpoint of CD.

Hence, AE = ED = DF = CF = 20

Let AB = c and BC = a

Applying Apollonius theorem in ∆ABC, we get,

BD2 + 402 = 1/2 × (c2 + a2) = 1/2 × 802

BD2 = 402 … (i)

Now, applying Apollonius theorem in ∆ABD, we get,

BE2 + 202 = 1/2 × (c2 + BD2) … (ii)

Similarly, applying Apollonius in ∆CDB, we get,

BF2 + 202 = 1/2 × (a2 + BD2) … (iii)

Adding (ii) and (iii), and substituting value from (i), we get,

BE2 + BF2 + 2 × 202 = 1/2 (a2 + c2 + 2 × BD2) = 1/2 (802 + 2 × 402) = 3 × 402

Hence, BE2 + BF2 + BD2 = 3 × 402 + 402 – 2 × 202 = 5600

Hence, option (c).

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