The sides of the triangle are $a, b, c$. Find the distance between the projections. The sides of the triangle are $\space a, b, c. \space$  Find the distance between the projections of the vertex onto the bisectors of the external angles at the other two vertices.
I've got that there are two right triangles that create the quadrangles. Then this quadrangle can describe a circle and it turns out that there some angles are equal but it didn’t give me anything.
 A: 
Here $BF$ and $CG$ are the external angle bisectors of $\angle ABC=\angle B$ and $\angle ACB=\angle C$, so $\angle ABF = \dfrac12 (\pi - \angle B) = \dfrac{\pi}2-\dfrac{\angle B}2$ and $\angle ACG = \dfrac{\pi}2 - \dfrac{\angle C}2$.
$\Delta AFB, \Delta AGC$ being right angled triangles, we have $\angle BAF=\dfrac{\angle B}2, \ \angle CAG = \dfrac{\angle C}2$, $$AB\sin(\angle ABF) = AF , \ AC\sin(\angle ACG) = AG \\ AF = c\sin\left(\dfrac{\pi}2-\dfrac{\angle B}2\right) = c\cos\left(\dfrac{\angle B}2\right), \ AG =b\cos\left(\dfrac{\angle C}2\right) $$
and $\angle FAG = \angle FAB + \angle BAC + \angle CAG = \dfrac{\angle B}2 +\angle A + \dfrac{\angle C}2 = \dfrac{\angle A}2 + \dfrac{\angle A +\angle B+\angle C}2 = \dfrac{\angle A}2+\dfrac{\pi}2$
Having $$AF=c\cos\left(\dfrac{\angle B}2\right), AG=b\cos\left(\dfrac{\angle C}2\right), \angle FAG = \dfrac{\angle A}2+\dfrac{\pi}2$$ we can use the half-angle formulae in terms of sides which gives $$AF=c\sqrt{\dfrac{s(s-b)}{ac}}, AG=b\sqrt{\dfrac{s(s-c)}{ab}}, \ \text{ where } s=\dfrac{a+b+c}2$$ and apply the cosine rule in $\Delta AFG$ to find the length of $FG$.
Incidentally, the cosine rule calculations are not that bad: $$FG^2=b^2\dfrac{s(s-c)}{ab} + c^2\dfrac{s(s-b)}{ac} -2bc\sqrt{\dfrac{s(s-c)}{ab}}\sqrt{\dfrac{s(s-b)}{ac}}\cos\left(\dfrac{\angle A}2 +\dfrac{\pi}2\right) \\= \dfrac{s}{a}(sb-bc+sc-bc) + \dfrac{2bcs}{a}\sqrt{\dfrac{(s-b)(s-c)}{bc}}\sin\left(\dfrac{\angle A}2\right)  \\ = \dfrac{s}a \left(sb+sc-2bc +2bc\sqrt{\dfrac{(s-b)(s-c)}{bc}}\sqrt{\dfrac{(s-b)(s-c)}{bc}}\right) \\ = \dfrac{s}a(sb+sc-2bc +2(s-b)(s-c)) = \dfrac{s}a(2s^2-(b+c)s)=\dfrac{s^2}a(2s-(b+c))\\ =\dfrac{s^2a}a=s^2 \implies FG = s = \dfrac{a+b+c}2$$
This now makes me feel that there can be nice ways of arriving at this length. For a purely geometric approach, my mind thinks of splitters and cleavers at the moment.
