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It is known that if point $D$ is the orthocenter [the uniquely determined point of intersection of all altitudes] of triangle $ABC$ then the following holds: $$BD\cdot DE = FD\cdot DA = GD\cdot DC \tag{1}$$ Now, let $GC$, $AF$ and $EB$ be some lines that intersect at common point $D$ and let $(1)$ also be satisfied. Is it sufficient for $D$ to be the orthocenter? Equivalently, are the lines through $A, B, C$ perpendicular to their respective opposite sides? enter image description here

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It helps order things in the mind if the naming is consistent -- starting ordering with the triangle vertex we are given

$AD \cdot DF = BD \cdot DE = CD \cdot DG$ in triangle $ABC$.

Doing some dividing, we see $AD/DE = BD/DF; AD/DG = CD/DF ; BD/DG = CD/DE$.

Then we have a bunch of vertically opposite angles.

$ \angle ADE = \angle BDF,\ \ \angle ADG = \angle CDF,\ \ \angle BDG = \angle CDE$.

So we have a lot of similar triangles.

$ \triangle ADE \sim \triangle BDF,\ \ \triangle ADG \sim \triangle CDF,\ \ \triangle BDG \sim \triangle CDE$.

Then we have a lot of equal corresponding angles

$\angle AGD = \angle CFD,\ \ \angle AED = \angle BFD,\ \ \angle BGD = \angle CED$

PLEASE NOTE: Correction: I made an unjustified assumption and hurried to the conclusion here. Please OMIT the part below between the ********* lines, and instead as in comments below work with supplementary angles and an easy chain of equalities:

After proving $ \angle AGD= \angle CFD, \angle AED=\angle BFD, \angle BGD=\angle CED $

Then as supplementary angles $ \angle AED +\angle CED=180; \angle BGD+\angle AGD=180; \angle CFD+\angle BFD=180. $

$\angle BFD = \angle AED =180 -\angle CED = 180 - \angle BGD = \angle AGD = \angle CFD $ and BFC is a straight line so $\angle BFD = \angle CFD = 90$ And similarly for the other two sides of the triangle.


OMIT below

Pairs of these are opposite angles in quadrilaterals so must sum to 180 degrees.

$ \angle AED + \angle AGD = 180; \ \ \angle BGD + \angle BFD = 180; \ \ \angle CED + \angle CFD = 180.$

Suppose $\angle ADE = \angle BFD < 90$ Then by the above sums $\angle AGD = \angle BGD > 90$ which is impossible since AGB is a straight line. Similarly these angles cannot be $>90$. We can work similarly around all three sides of the triangle.

OMIT above


$\angle AGD = \angle CFD = \angle AED = \angle BFD = \angle BGD = \angle CED = 90$

and $AF, BE, CG$ are all altitudes as required.

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  • $\begingroup$ Thank you for your answer. We can also arrive at desired result directly $\alpha :=\angle AGD = \angle CFD,\ \ \beta :=\angle AED = \angle BFD,\ \ \gamma :=\angle BGD = \angle CED$. We know $\gamma + \beta = \gamma+\alpha =\alpha+\beta =\pi$. It follows that $\alpha = \beta = \gamma =\pi /2$ $\endgroup$ – Alvin Lepik Feb 26 '17 at 11:04
  • $\begingroup$ Well yes, but first how do you know $\alpha = \angle AGD = \angle CFD$ etc.? $\endgroup$ – victoria Feb 27 '17 at 18:58
  • $\begingroup$ I was thinking of an alternative conclusion also. After proving $\endgroup$ – victoria Feb 27 '17 at 19:00
  • $\begingroup$ Sorry, timed out. After proving ∠AGD=∠CFD, ∠AED=∠BFD, ∠BGD=∠CED and ∠AED+∠AGD=180; ∠BGD+∠BFD=180; ∠CED+∠CFD=180. Then $\angle BFD + \angle CFD = 180$ So with a little add and subtract and substitution we get $\angle BFD = \angle CFD = 90.$ $\endgroup$ – victoria Feb 27 '17 at 19:11
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    $\begingroup$ @AlvinLepik There is one important detail in this proof that is not quite right. In general, pairs of opposite angles in quadrilaterals do not sum up to 180. It happens if and only if the quadrilaterals are inscribed in circles. one has to prove that. $\endgroup$ – Futurologist Feb 28 '17 at 15:47
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The identity $BD \cdot DE = FD \cdot DA$ holds if and only if the four points $A, B, F$ and $E$ lie on a common circle, call it $g$. Analogously, $FD \cdot DA = CD \cdot DC$ holds if and only if the four points $A, C, F$ and $G$ lie on a common circle, call it $e$. By the same argument the points $B, C, E$ and $G$ lie on a common circle $f$.

Observe that in circle $e$ the following angle equality holds $$\angle \, FCG = \angle \, FAG = \angle \, FAB$$ In circle $g$ the following angle equality holds $$\angle \, FEB = \angle \, FAB$$ Therefore $$\angle \, FCD = \angle \, FCG = \angle \, FEB = \angle \, FED$$ Therefore the quad $CEDF$ is inscribed in a circle so $$\angle \, DEC = 180^{\circ} - \angle \, DFC = \angle \, BFD = \angle \, BFA$$ Hence $$\angle \, BEC = \angle \, DEC = \angle \, BFA$$ However in circle $g$ $$\angle \, BEA = \angle \, BFA$$ so $\angle \, BEA = \angle \, BFA = \angle \, BEC$ and since $E$ lies on the line $AC$, the identity $\angle \, BEA = \angle \, BEC$ implies that $BE$ is orthogonal to $AC$.

Analogously, one shows that $AF$ is orthogonal to $BC$ and $CG$ is orthogonal to $AB$.

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