If triangle $DEF$ is the orthic triangle of triangle $ABC$, the symmedian points of triangles $AEF$, $BFD$, $CDE$ lie on the medians of $ABC$

Question

If $\triangle DEF$ is the orthic triangle of $\triangle ABC$, show that the symmedian points of the triangles $\triangle AEF$, $\triangle BFD$, $\triangle CDE$ lie on the medians of $\triangle ABC$.

I found this in textbook called "College Geometry An Introduction to the Modern Geometry of the Triangle and the Circle", but the solution wasn't shown there. I only know that $\triangle AEF$, $\triangle BFD$, $\triangle CDE$, and $\triangle ABC$ are similar, but I don't know what to do next.

Answer 1

$\mathrm{Fig.\space 1}$ shows the complete scenario described by your question. $\mathrm{Fig. \space 1}$ is reproduced in $\mathrm{Fig. \space 2}$ with all those lines unnecessary for our proof erased. We work out the problem in three different steps to establish the proof we seek.

$\mathrm{\mathbf{\underline {Step\space 1}:}}$ Proving that the corner triangles $AH_CH_B$, $BH_AH_C$, and $CH_BH_A$ are inversely similar to the triangle $ABC$.

Let $\measuredangle CAB = \alpha$, $\measuredangle ABC = \beta$, and $\measuredangle BCA = \phi$.

Consider the two right-angle triangles $BCH_C$ and $BCH_B$. They have a common hypotenuse $BC$. Therefore, $BCH_BH_C$ is a cyclic quadrilateral. As a consequence of this, $\measuredangle CH_BH_C = 180^0 – \beta$. Since $\measuredangle H_CH_BA$ is the supplement of $\measuredangle CH_BH_C$, $$\measuredangle H_CH_BA = \beta = ABC.$$

In like manner, we can show $\measuredangle AH_CH_B = BCA$. Therefore, all three corresponding angles of the triangles $ABC$ and $AH_CH_B$ are equal, but described in the opposite rotational sense. Hence, they are inversely similar.

Similar reasoning yields that $BH_AH_C$ and $CH_BH_A$ are also inversely similar to $ABC$.

$\mathrm{\mathbf{\underline{Step\space 2}:}}$ Proving that $BP$ is one of the medians of the corner triangle $BH_AH_C$

Note that $BM_B$ is one of the medians of the triangle $ABC$.

Now, pay attention to the following construction of a pair of lines. First, the angle bisector $BQ$ of the angle $ABC$ is drawn to meet $AC$ at $Q$. Then, the line $BP$ is drawn to meet $H_CH_A$ at $P$, so that it and $BM_B$ make equal angles (i.e. $\theta$) with $BQ$. $BP$ is called the isogonal line of $BM_B$ and vice versa.

Consider the triangles $BCM_B$ and $BPH_C$. It is obvious from $\mathrm{Step\space 1}$ that $\measuredangle BCM_B = \measuredangle PH_CB$. According to the construction, $\measuredangle M_BBC = \measuredangle H_CBP$. Therefore, the two triangles are inversely similar. Likewise, the two triangles $ABM_B$ and $PBH_A$ are also inversely similar. Now, we can write for the former pair of inversely similar triangles, $$\frac{CM_B}{PH_C}=\frac{BM_B}{BP}$$ and for the later pair of inversely similar triangles, $$\frac{AM_B}{PH_A}=\frac{BM_B}{BP}.$$ $$\therefore\space\space \frac{CM_B}{PH_C}=\frac{AM_B}{PH_A}.\space\space\space\space\space$$ Since $ CM_B = AM_B $, $ PH_C = PH_A $. This means that $BP$ is a medians of the corner triangle $BH_AH_C$.

$\mathrm{\mathbf{\underline{Step\space 3}:}}$ Proving that the symmedian points of the corner triangles lie on the medians of the triangle $ABC$

We make use of the fact that the symmedian point of a triangle is the isogonal conjugate of its centroid and vice versa. This means that the median passing through a certain vertex of a triangle and the line joining its symmedian point to the same vertex are isogonal lines.

In our case, the centroid of the corner triangle $BH_AH_C$ lies on $BP$, which is a median (proved in $\mathrm{Step\space 2}$). Then, the symmedian point of the corner triangle $BH_AH_C$ must lie on the isogonal line of $BP$, which happened to be $BM_B$. By the way, $BM_B$ is a median of the triangle $ABC$.

It is possible to prove that the other two symmedian points lie on the medians of the triangle $ABC$ by using a similar sequence of arguments.

Answer 2

I am trying to give a compact, but still detailed proof. Let us fix the notation as in the following picture:

$\Delta DEF$ is the ortic triangle, $A'$ is the mid point of $BC$, $S$ is the mid point of $EF$.

Proof of the property claimed in the OP:

• $BCEF$ is cylic because of the right angles in $E,F$. This implies $\widehat{AFE}=\widehat{ECB}$ and $\widehat{AEF}=\widehat{FBC}$.

• We obtain the similarity of triangles $\Delta ABC\sim\Delta AEF$ (with these orders of vertices respectively). This gives the proportionality $$ \color{blue}{ \frac{AF}{AC}} = \frac{AE}{AB} = \frac{EF}{BC} = \frac{EF/2}{BC/2} = \color{blue}{ \frac{FS}{A'C} } \ . $$

• The equality of the blue proportions together with the equality of angles in $F$ and $C$ in the marked triangles $\Delta AFS$ and $ACA'$ show their similarity. In particular the angles in $A$ in the two triangles are equal, so reflecting w.r.t. the angle bisector of $\widehat{BAC}$ the line $AS$ (supporting the median of $\Delta AEF$) we obtain the line $AA'$ (supporting the median of $\Delta ABC$).

• So the line $AA'$ supports the symmedian of $\Delta AEF$, and the median of $\Delta ABC$. We have similar properties for $BB'$, $CC'$ (using similar notations) instead of $AA'$.

$\square$