I was solving this question. I solved it. Then I forgot how I solved it.

An acute angled triangle $ABC$ has $AD$ as altitude. $E$ is the midpoint of $BC$. $F$ is the midpoint of $AC$. $\angle{EAB}=40°$ and $\angle{EAD}=\angle{EFD}$. Find $\angle{ADF}$.

This is the diagram I made. Made by a Pen

What I cannot understand is how I proved $\angle{AEF}=\angle{ADF}$, or if that is even true, and how to actually get the solution.

Can anyone help?


The points $A$, $F$, $E$, and $D$ are concyclic because $A$ and $F$ subtends equal angles with respect to the segment $ED$ that is $\angle{EAD}=\angle{EFD}$. Moreover $EF$ is parallel to $AB$ because $\triangle ABC$ and $\triangle FEC$ are similar. Hence $$40^{\circ}=\angle{EAB}=\angle{AEF}=\angle{ADF}$$ where in the last equality we used the fact that $A$, $F$, $E$, and $D$ are concyclic and therefore $E$ and $D$ subtends equal angles with respect to the segment $AF$.

  • $\begingroup$ I knew that... But how does that help? $\endgroup$ – MalayTheDynamo Jul 10 '17 at 14:05
  • $\begingroup$ @MalayTheDynamo I edited my answer with more details. $\endgroup$ – Robert Z Jul 10 '17 at 14:13
  • $\begingroup$ Fuu says thanks. Tobi is a good boy. $\endgroup$ – MalayTheDynamo Jul 10 '17 at 14:18

$40^{\circ}$ because $AEDF$ is cyclic and $EF||AB$.

  • $\begingroup$ Please elaborate. $\endgroup$ – MalayTheDynamo Jul 10 '17 at 14:04
  • $\begingroup$ @MalayTheDynamo $E$ and $F$ are midpoints of $BC$ and $AC$ respectively. Thus, $EF||AB$. $\endgroup$ – Michael Rozenberg Jul 10 '17 at 14:15
  • $\begingroup$ No, I know the midpoint theorem. But how do the parallelism(?) of the lines and a concyclic quadrilateral come together to give 40°? $\endgroup$ – MalayTheDynamo Jul 10 '17 at 14:16
  • $\begingroup$ @MalayTheDynamo They are alternate angles. $\endgroup$ – Michael Rozenberg Jul 10 '17 at 14:18
  • $\begingroup$ @Michael Rozenberg (+1) For the hints. $\endgroup$ – Robert Z Jul 10 '17 at 14:21

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