In triangle $ABC$, $AB = AC$, $D$ is the midpoint of $\overline{BC}$, $E$ is the foot of the perpendicular from $D$ to $\overline{AC}$, and $F$ is the midpoint of $\overline{DE}$. Prove that $\overline{AF}$ is perpendicular to $\overline{BE}$.

My first approach was to align the triangle in the first quadrant, on the x-coordinates and started calculating slopes and positions of points. But then things got messy real fast, i'm afraid I'm approaching this problem the wrong way. Is there a better way? No trigonometry just yet! Solutions are greatly appreciated. Thanks in advance!

  • $\begingroup$ Are you required an analytic solution? $\endgroup$ – Stefan4024 Nov 2 '16 at 20:27
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    $\begingroup$ Questions about geometry, any geometry, are greately enhanced by adding a diagram. $\endgroup$ – DonAntonio Nov 2 '16 at 20:34
  • $\begingroup$ @Stefan4024 An analytic solution would be best. $\endgroup$ – Dreamer Nov 2 '16 at 20:46

But maybe if you saw the geometric proof, you would change your mind about looking for an analytic proof of this fact.

Take $K$ to be the midpoint of segment $CE$. Then $KF$ is a mid-segment in triangle $DCE$ and therefore $KF$ is parallel to $CD$. Furthermore, $CD$ is orthogonal to $AD$, which means that line $KF$ is orthogonal to $AD$ too. Hence the lines $DE$ and $KF$ are altitudes in triangle $DKA$ and consequently point $F$ is the orthocenter of $DKA$. Therefore, the line $AF$ is the third altitude in triangle $DKA$ and so $AF$ is orthogonal to $DK$. The segment $DK$ however is mid-segment of triangle $BCE$ so $DK$ is parallel to $BE$. Thus $BE$ is orthogonal to $AF$.

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    $\begingroup$ Pure beauty! +1 $\endgroup$ – Qwerty Nov 3 '16 at 5:42

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We may embed the construction in the complex plane, assuming $D=0$, $C=1, B=-1$ and $A=ri$. With such assumptions we may compute $E$, then $F=\frac{E}{2}$ and check that $\frac{F-A}{E-B}$ is a purely imaginary number, proving $AF\perp BE$.

  • $\begingroup$ [+1] Very short solution... My one uses another tool : vector geometry... $\endgroup$ – Jean Marie Nov 2 '16 at 22:35

Here is a vectorial proof with a little trigonometry at the end. We can transform our objective thus:

$$\tag{2}AF \perp BE \ (a) \ \ \ \ \iff \ \ \ 2\vec{AF} \cdot \vec{BE}=0 \ (b)$$

(symbol $\cdot$ meaning "dot product"). The LHS of (2b) can be transformed thus:

$$(\vec{AD}+\vec{AE}) \cdot (\vec{BD}+\vec{DE})=\vec{AD}\cdot\vec{DE} + \vec{AE} \cdot \vec{BD} $$

(taking into account the orthogonality of AD and BD, and AE and DE).

Let $\alpha=\widehat{BAC}=\widehat{DAC}$ and $h$ be the length of altitude $AD$.

It is easy to establish that:

$$\begin{cases} \vec{AD} \cdot \vec{DE}&=&- h \times h \sin(\alpha) & \ \ \text{(the minus sign is essential)}\\ \vec{AE} \cdot \vec{BD} &=& h \cos(\alpha) \times h \tan(\alpha) & \end{cases}$$

(where $\times$ is the ordinary multiplication of real numbers).

Adding these two results gives $0$ ; we have thus obtained the RHS of (2).

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    $\begingroup$ +1 for vectorial, and a variation without the trig: $$ \require{cancel} \begin{align} 2 \;\vec{AF} \cdot \vec{BE} & \;= \; \left(\vec{AE}+\vec{AC}+\vec{CD}\right) \cdot \left(-\vec{CD}+\vec{DE} \right) \\ & \;=\; -\vec{AE}\cdot \vec{CD} + \cancel{\vec{AE} \cdot \vec{DE}} -\vec{AC}\cdot\vec{CD} + \bcancel{\vec{AC}\cdot\vec{DE}} - \vec{CD}\cdot\vec{CD}+\vec{CD}\cdot\vec{DE} \\ & \;=\; -\vec{CD} \cdot \left(\big(\vec{AE}-\vec{DE}\big)+\big(\vec{AC}+\vec{CD}\big) \right) \;=\; -2 \; \vec{CD}\cdot\vec{AD} \;=\; 0 \end{align} $$ $\endgroup$ – dxiv Nov 2 '16 at 23:08
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    $\begingroup$ @dxiv congratulations for this witty decomposition ! $\endgroup$ – Jean Marie Nov 2 '16 at 23:25

I know a coordinate geometric proof will hurt the true lovers of geometry, but still I would like to present one as the OP found it MESSY.

Without loss of generality, position the triangle as $A=(0,0)\ B=(-1,k)\ C=(1,k)$

Then check that $D=(0,k)$

Let $E$ be $(a,ka):a\in(0,1)$

By the problem , $${ka-k\over a}\cdot k=-1\implies a={k^2\over k^2+1}$$

So $E=\left({k^2\over k^2+1},{k^3\over k^2+1}\right)$

Another simple calculation reveals $F$ to be $\left({k^2\over 2(k^2+1)},{2k^3+k\over 2(k^2+1)}\right)$

Now its trivial that $$\text{Slope}_{BE}={{k^3\over k^2+1}-k\over {k^2\over k^2+1}+1 }=-{k\over 2k^2+1}={-1\over \text{Slope}_{AF}}$$

  • $\begingroup$ Sorry if I offended you in any way, I was just stating that my way was probably incorrect because it was nice one moment then huge equations the next. $\endgroup$ – Dreamer Nov 2 '16 at 21:03
  • $\begingroup$ @Regina No offense taken. Cheers! Just remember that one messy looking equation doesn't mean that the entire solution will be messy. $\endgroup$ – Qwerty Nov 2 '16 at 21:05
  • $\begingroup$ Analytic geometry proofs should never hurt true lovers of geometry [+1]. $\endgroup$ – Jean Marie Nov 2 '16 at 21:18

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