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Many years back in high school I happened to stumble upon the following property that seems to hold for any acute triangle:

enter image description here

$CD$ and $BE$ are altitudes, the arcs are semicircles with diameters $AB$ and $AC$ respectively.

The property is that $AF = AG$

Proof:

Let $H$ be the midpoint of $AB$ (and the centre of the respective semicircle) $$AG^2 = AD^2 + GD^2 = \left(AC\cdot \cos\angle A\right)^2 + GD^2$$ Since $HG = AH = \frac{AB}{2}$ is the radius of the semicircle $$GD^2 = HG^2 - HD^2 = \left(\frac{AB}{2}\right)^2 - \left(\frac{AB}{2} - AC\cdot\cos\angle A\right)^2 = \\ = AB\cdot AC\cdot \cos\angle A - \left(AC\cdot\cos\angle A\right)^2$$

which gives $$AG^2 = AB\cdot AC\cdot \cos\angle A$$

enter image description here

Analogously ($I$ is the midpoint of $AC$) $$AF^2 = AE^2 + FE^2 = \left(AB\cdot \cos\angle A\right)^2 + FE^2$$ $$FE^2 = FI^2 - EI^2 = \left(\frac{AC}{2}\right)^2 - \left(AB\cdot \cos\angle A - \frac{AC}{2}\right)^2 = \\ = AC\cdot AB\cdot \cos\angle A - \left(AB\cdot \cos\angle A\right)^2$$ which finally gives $$AF^2 = AC\cdot AB\cdot \cos\angle A$$ enter image description here

Questions:

  1. Is this a known property?
  2. Is there a better more elegant proof?
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    $\begingroup$ You can say something more: The four points where the (full) circles with diameters $\overline{AB}$ and $\overline{AC}$ meet the altitudes from $B$ and $C$, lie on a common circle about $A$. $\endgroup$
    – Blue
    Apr 3 '18 at 16:49
  • $\begingroup$ @Blue, that's a very, very good one! Thank you. $\endgroup$
    – ayorgo
    Apr 3 '18 at 21:54
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HINT:

$$AF^2 = AE \cdot AC\\ AG^2 = AD \cdot AB \\ AE \cdot AC = AD \cdot AB $$

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    $\begingroup$ what a beautiful proof! $\endgroup$
    – CY Aries
    Apr 3 '18 at 16:42
  • $\begingroup$ Brilliant indeed. Although not so trivial (to me at least). Thanks a lot. $\endgroup$
    – ayorgo
    Apr 3 '18 at 21:41
  • $\begingroup$ @ageorge: My pleasure! A nice problem indeed! $\endgroup$
    – orangeskid
    Apr 3 '18 at 22:32
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This is a possibly similar proof, but here goes anyway...

Let the triangle be labelled in the usual way, so $AC=b$ and $AB=c$, and let angle $FAC=\theta$

Then $$AF=b\cos\theta$$ $$\implies FE=AF\sin\theta=b\sin\theta\cos\theta$$ $$\implies EC=FE\tan\theta=b\sin^2\theta$$ But $$EC=b-c\cos A=b\sin^2\theta$$ $$\implies b\cos^2\theta=c\cos A$$ $$\implies AF=\sqrt{bc \cos A}$$

To get $AG$ we only need to exchange $b$ and $c$, so the result follows.

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