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Explanation

We know that if in `Delta ABC, AD` divides BC in the ratio `m : n` with `angle ADC = theta, angle BAD= alpha and angleCAD = beta`

The m-n theorem states that
`(m + n) cot theta = m cot alpha - n cot beta` (i)
By applying Eq. (i) in `Delta ABC`, we get
`(1 + 1) cot theta = 1 cot 30^(@) - 1 cot 45^(@)`
or `2 cot theta = sqrt3 - 1`
or `cot theta = (sqrt3 -1)/(2)`
or `tan theta = (2)/(sqrt(3 -1))`
But `theta = B + 30^(@)`, we get
`tan (B + 30^(@)) = (2)/(sqrt3 - 1) = (2(sqrt3 + 1))/(3 - 1) = sqrt + 1`
or `(tan B + (1)/(sqrt3))/(1 - (1)/(sqrt3) tan B) = sqrt3 + 1`
or `(sqrt3 tan B + 1)/(sqrt3 - tan B) = sqrt3 + 1`
or `sqrt3 tan B + 1 = 3 + sqrt3 - (sqrt3 + 1) tan B`
or `(2sqrt3 + 1) tan B = 2 + sqrt3`
or `tan B = (2 + sqrt3)/(2 sqrt3 + 1)`
or `cot B = (2 sqrt3 + 1)/(2 + sqrt3) xx (2 - sqrt3)/(2 - sqrt3) = (4 sqrt3 - 6 + 2 - sqrt3)/(4 - 3) = 3 sqrt3 - 4`
or `cosec^(2) B = 1 + 27 + 16 - 24 sqrt3 = 44 - 24 sqrt3`
or `sin^(2) B = (1)/(4 (11 - 6 sqrt3))`
or `sin B = (1)/(2 sqrt(11 - 6 sqrt3))` (i)
Now in `Delta ABD`, by applying the since law, we get
`(BD)/(sin 30^(@)) = (AD)/(sin B)`
or `BD = (AD)/(sin B) xx sin 30^(@) = ((1)/(sqrt(11 - 6 sqrt3)))/((1)/(2sqrt(11 - 6 sqrt3))) xx (1)/(2) = 1`
`:. BC = 2BD = 2` units