# how to prove that the segment $IF=HF+GF$

$$AE$$ and $$CD$$ are the angle bisectors of $$\triangle ABC$$. $$F$$ is an arbitrary point on line $$DE$$. Prove that $$GF+HF=IF$$.

I noticed $$3$$ cyclic quadrilaterals. Any ideas. Here is the picture

• I don't agree to close a puzzling question where the asker has taken time to draw a nice figure ; Aug 16, 2020 at 7:14

Consider Trilinear Coordinates (https://en.wikipedia.org/wiki/Trilinear_coordinates) first in the case where $$F$$ is inside triangle $$ABC$$.

$$D$$ and $$E$$, being feet of angle bissectors, have resp. trilinear coord. $$(1,1,0)$$ and $$(0,1,1)$$. Therefore, the trilinear equation of straight line $$DE$$ is:

$$\begin{vmatrix}1&0&x\\1&1&y\\0&1&z\end{vmatrix}=0 \ \ \iff \ \ x-y+z=0\tag{0}$$

Interpreting $$(x=FG,y=FH,z=FI)$$, we get:

$$FG+FI-FH=0\tag{1}$$

(which is not the given relationship !)

Now, if $$F$$ is not inside triangle $$ABC$$, here are the other cases:

• In the case depicted in the given figure ($$F$$ "just outside" $$[DE]$$ on the side of $$E$$), only one of the trilinear coordinates, $$FG$$, undergoes a sign change ; therefore (1) becomes:

$$\color{red}{-}FG+FI-FH=0\tag{2}$$

which amounts to the given relationship, this time !

If, in the case of the given figure, $$F$$ is far away, a second sign change occurs, now for signed distance $$FH$$, transforming (2) into :

$$-FG+FI\color{red}{+}FH=0\tag{3}$$

which is a third formula.

• if, on the contrary, $$F$$ is outside of line segment $$[D,E]$$ but on the side of $$D$$, we have to change $$FI$$ into its opposite in (1), giving back relationship (3).

Remark about relationship (0): we have obtained it by working up to a multiplicative constant ; this is unimportant because we deal with relationships having a zero in their right-hand side.

• Could you explain to me how the relation and the how matrix will change when $F$ is inside the segment $DE$ and to the left side of $DE$? It seems some negative signs need to be added but I am confused about the rule of the signs. Aug 16, 2020 at 8:18
• The case where $F$ is inside line segment [DE] is without problem, because trilinear coordinates are all positive in this case, therefore are the same as (positive) distances to the sides. For the other cases, I have to add an Edit to my answer. It will be done shortly Aug 16, 2020 at 8:28
• In the process of writing this edit, I realized that the relationship you give is valid only on a certain portion of line (DE), but must be replaced by other (cousin) relationships for other parts of the line... This explains why you were confused about the sign management... Aug 16, 2020 at 11:57
• Thank you for the clarification. It all makes sense to me now. Aug 16, 2020 at 11:59
• Could you tell the origin of the problem ? Aug 21, 2020 at 20:47