Let the perpendiculars from A to BF and from C to BD intersect at O.

Let the position vectors of points A, B, ...relative to O, be a, b, ... with magnitudes a, b, ...

Let M be the mid-point of AC, with position vector 0.5(a + c).

AB = BC, so MB is perpendicular to AC, giving[b - 0.5(a + c)].(c - a) = 0

This can be simplified tob.c - b.a = 0.5(c^{2} - a^{2})

Using the other two sides in a similar way givesd.e - d.c = 0.5(e^{2} - c^{2})

andf.a - f.e = 0.5(a^{2} - e^{2})

Adding these three equation:f.a - b.a + b.c - d.c + d.e - f.e = 0

a.(f - b) + c.(b - d) + e.(d - f) = 0

Since OA is perpendicular to BF and OC is perpendicular to BD, the first two terms are both zero. This leaves e.(d - f) = 0, proving that OE is perpendicular to DF and therefore that the three perpendiculars all pass through the same point, O.

This property seems to be independent of the sum of angles B, D and F, so the constraint that B + D + F = 300 degrees seems unnecessary (or have I missed something?).