4
$\begingroup$

I need a synthetic proof on this problem without the use of trigonometry.

enter image description here
Question:
Let $ABC$ be a triangle with $AB=AC$. If $D$ is the midpoint of $BC$, $E$ the foot of perpendicular drawn from $D$ to $AC$ and $F$ the midpoint of $DE$, prove that $AF\perp BE$.

The proof is very easy using coordinate geometry.

I'm stuck at: Since point $F$ is midpoint of $DE$, it is fixed and is not a trivial information. But I can't see a way to use this information.

$\endgroup$
5
$\begingroup$

Let $G$ be a midpoint for $CE$. Then $DG$ is a middle line in triangle $BCE$ so $DG||BE$. Also we see that $FG||DC$ so $FG\bot AD$. So $F$ is an orthocenter in triangle $ADG$. So ...

$\endgroup$
  • $\begingroup$ Nice solution! +1 $\endgroup$ – Michael Rozenberg Aug 1 '18 at 11:48
  • $\begingroup$ Really nice solution. $\endgroup$ – Love Invariants Aug 1 '18 at 16:05
2
$\begingroup$

Let $K$ be a midpoint of $BD$.

Thus, since $$\Delta ABD\sim\Delta DCE,$$ we obtain: $$\frac{AB}{DC}=\frac{BD}{CE}$$ or $$\frac{2AB}{BC}=\frac{2BK}{CE}$$ or $$\frac{AB}{BC}=\frac{BK}{CE}$$ and since $\measuredangle ABK=\measuredangle BCE,$ we obtain $$\Delta ABK\sim\Delta BCE,$$ which gives $$\measuredangle BKA=\measuredangle BEC.$$ Now, let $BE\cap AK=\{L\}.$

Thus, $$\measuredangle BLA=\measuredangle LAE+\measuredangle BEA=\measuredangle KAD+180^{\circ}-\measuredangle BEC=$$ $$=\measuredangle KAD+180^{\circ}-\measuredangle BKA=180^{\circ}-90^{\circ}=90^{\circ}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.