Prove that the sides of the orthic triangle meet the sides of the given triangle in three collinear points.

Question

Prove that the sides of the orthic triangle($DEF$) meet the sides of the given triangle($ABC$) in three collinear points($X, Y, Z$) .

Prove using Menelaus theorem, Stewart's theorem, ceva's theorem

I tried to find the ratios $\frac{YA}{YB}$, $\frac{XA}{XC}$ and $\frac{ZB}{ZC}$ by getting individual values of these lengths via calculating some lengths by pythagoras and some by stewarts theorem. But I ended up with some complicated equations which when I tried to solve I was getting stuck in variables.

Answer 1

As you have mentioned menelaus, i would like to post the proof using cross ratios without defining cross ratios (means a little,easy exercise for you). I will avoid directed lengths.

It is well known that $ (XE,AC)=-1$ or $\frac{CE}{CX}\frac{AX}{AE}=1$ (show this!). Similiarly, $(BA,FY)=-1$ or $\frac{YA}{YB}\frac{FB}{FA}=1$ (show this!) and finally, $(CB,DZ)=-1$ or $\frac{ZB}{ZC}\frac{DC}{DB}=1$ (show this!). Now multiply all these to get the resukt immediately.

Answer 2

This is just Desargues Theorem, but i will give an elementary proof for this one using Radical Axes and stuffs. Basically if you are taking the Menelaus approach, you solve this problem when you actually prove Desargues Theorem using only Menelaus.

It is trivial to note that $AFDC$ is cyclic as $\measuredangle AFC=\measuredangle ADC=90$. Draw the circumcircle of $\Delta ABC$ and note that by Power of A Point we havve $\text{Pow}_{\odot (AFDC)}=XF\cdot XD = XA\cdot XC$ ($AFDC$ is cyclic) $=\text{Pow}_{\odot (ABC)}$ implying that $X$ lies on the radical axis of the circles $\odot (ABC), \odot (AFDC)$. Using a similar argument, $Y,Z$ lie on the radical axis of these circles as well and we are done.

(There are solutions using poles/polars, harmonic bundles too, for solving it with desargues, one needs to only show $AD,BE,CF$ are concurrent which is trivial)

Answer 3

By Ceva's Theorem on $\triangle ABC$ with respect to $O$(Orthocentre): $\frac{AE.CD.BF}{EC.DB.FA}=1$.....(1)

Menelaus Theorem on $ABC$,

with respect to $DEY$:$\frac{BD.CE.AY}{DC.EA.YB}=-1$.....(2)

with respect to $DFX$:$\frac{BD.CX.AF}{DC.XA.FB}=-1$.....(3)

with respect to $ZFE$:$\frac{BZ.CE.AF}{ZC.EA.FB}=-1$.....(4)

Combining (2), (3), (4) : $\frac{AY.CX.BZ}{YB.XA.ZC}=-\bigg (\frac{AE.CD.BF}{EC.DB.FA} \bigg)^2$.....(5)

Combining (5), (1) : $\frac{AY.CX.BZ}{YB.XA.ZC}=-(1)^2=-1$

Therefore by Converse of Menelaus Theorem $X, Y, Z$ must be collinear.