The proof results by applying the cosine theorem on triangles ABD, ADC.

A question that may arise is if there exists a ratio r/s, such that all corresponding segments AD, BE, CF are equal.

Setting this common length AD

Considering this as a linear system in r

Two instances of the formula are well known: Applying this formula for the median (r=s=1), gives:

m

And for the internal bisector of the angle A (r = c, s = b):

u

v

For points D = (1-t)B+tC, BD/DC = t/(1-t) = r/s with r+s=1. The above formula becomes:

AD

BD/DC = t/(1-t) = (a

For a fixed t, the corresponding lengths AD

Eliminating t from the equations we find two equations determining AD

See the file SquaresCombination.html for a nice application of the theorem of Stewart.

Projection_Polygon.html

SquaresCombination.html

WeightedSumsOfDistances.html

Produced with EucliDraw© |