$AE$ 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 pictureenter image description here

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    $\begingroup$ I don't agree to close a puzzling question where the asker has taken time to draw a nice figure ; $\endgroup$
    – Jean Marie
    Aug 16, 2020 at 7:14

1 Answer 1


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:


(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:


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 :


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.

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    $\begingroup$ 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. $\endgroup$
    – cr001
    Aug 16, 2020 at 8:18
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    $\begingroup$ 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 $\endgroup$
    – Jean Marie
    Aug 16, 2020 at 8:28
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    $\begingroup$ 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... $\endgroup$
    – Jean Marie
    Aug 16, 2020 at 11:57
  • $\begingroup$ Thank you for the clarification. It all makes sense to me now. $\endgroup$
    – cr001
    Aug 16, 2020 at 11:59
  • $\begingroup$ Could you tell the origin of the problem ? $\endgroup$
    – Jean Marie
    Aug 21, 2020 at 20:47

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