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Let $ABC$ be a triangle, $F$ is a point inside the triangle such that $\angle ABF = \angle ACF$. $E$ and $D$ are the orthogonal projections of $F$ on $AB$ and $AC$, $G$ is the median of $BC$ , prove that $GD=GE$.

There is a problem with my diagram, I got that $E$ is the intersection of $(CF)$ with $AB$, same thing with D, is that just a construction error? And does proving it leads to proving the original statement?

The statement is equivalent to $\angle EDG=\angle DEG$, enter image description here

PNT
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  • There is a problem of your construction of the diagram. If C, E,F are collinear this implies that is the orthocenter. – JetfiRex Jan 24 '22 at 22:09
  • Please provide a better picture, instead of the misleading picture from above. Make sure there is a clear visual hint that $C,F,E$ are not collinear, and also that $B,F,D$ are not collinear. Please always show the own attempts to solve the problem. The source, the author, the level of the problem may best be also inserted, because providing context is one of the main chances to have a qualitatively good post, see also https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question. Also, JetfiRex has now a fairly simple, complete proof, consider giving credit to it. – dan_fulea Jan 29 '22 at 16:26

2 Answers2

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Take the $K$ and $L$ as the midpoint of $BF$ and $CF$. We have:

$\triangle BEF$ is rectangle triangle so $EK=FB/2$. Similarly, $DL=FC/2$.

$K$ is the midpoint of $FB$ and $G$ is the midpoint of $BC$, so $KG=FC/2$. Similarly, $LG=FB/2$.

$\angle EKF=2\angle EBF=2\angle DCF=2\angle DLF$. Also, $\angle FLG=\angle FKG$ since $FKGL$ is parallelogram. So $\angle EKG=\angle DLG$.

So, in the triangle $EKG$ and triangle $DLG$, we have $DL=GK$, $LG=KE$, $\angle DLG=\angle GKE$. So $EKG$ and $DLG$ are congruent. Therefore, $EG=DG$.

JetfiRex
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"$\angle BEC = \angle BDC = 90^0$" implies BCDE is cyclic with G as the center of that circle.

GD = GE because they are the radii of the same circle.

Mick
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  • There is a problem with the given picture, it is too special, what is claimed also allows non-collinear points $B,F,D$, and non-collinear points $C,F,E$. So the two angles are not given to be right angles... – dan_fulea Jan 28 '22 at 19:14
  • @dan_fulea (1)The question says "E and D are the orthogonal projections of F on AB and AC". (2) My answer is so short purely because "it is too special". – Mick Jan 29 '22 at 15:55
  • (1) $E$ is the orthogonal projection of $F$ on $AB$, yes, well maybe not also of $C$. So why do we have $\angle BEC=90^\circ$ (as the answer just starts)? (2) You are not giving any picture, the OP does it, and her/his picture is special, as this is also mentioned in the posted question. The picture is misleading, and it is easy to fall into its trap. This might also lead to very special short answers. Try to read the question again, the other answer, and also get the message in the comment of JetfiRex to the question. I will not put some -1 here, since i should do it also with the question. – dan_fulea Jan 29 '22 at 16:19
  • @dan_fulea You are right. I was misled by the OP's diagram. I will have my post deleted shortly. – Mick Jan 30 '22 at 04:49
  • This also happens to me very, very often, often it is my fault for reading in big hurry and really missing the problem, and solving my problem, often it is because of the post makes it very easy to "see an other problem" (with a simple solution). The latter is the case now. Please do not delete this post, just mention maybe as a later edit it is a fairly direct solution for the very special case highly suggested by the picture. Some eyes will see and understand the situation, this contributes certainly to the quality of next own posts. All the best! @Mick – dan_fulea Jan 30 '22 at 13:17
  • @dan_fulea Fine. Will leave it like that. – Mick Jan 30 '22 at 13:20