Feeds:
Bài viết
Bình luận

## Sondat’s theorem

Sondat’s theorem:

Given a triangle $ABC$. An arbitrary line d intersects $BC, CA, AB$ at $A', B', C'$, respectively. Construct the paralogic triangle $A_1B_1C_1$ of $ABC$ wrt $d$. Then $d$ bisects the line joining the two orthocenters $H$ and $H_1$ of $\Delta ABC$ and $A_1B_1C_1$.

Proof:

Lemma:  Given two similar triangles $ABC$ and $A_1B_1C_1$. Let $A_2, B_2, C_2$ be the midpoints of $AA_1, BB_1, CC_1$, respectively. Then $\Delta A_2B_2C_2\sim \Delta ABC \sim \Delta A_1B_1C_1$.
Proof:  Let $M, N$ be the midpoints of $AB_1$ and $AC_1$. It’s easy to see that $\Delta A_2MN\sim \Delta A_1B_1C_1$.
Use direct angle we can show that $(AB, A_1B_1)=(AC, A_1C_1)$, which follows that $\angle A_2MB_2=\angle A_2NC_2$. So $\Delta A_2MB_2= \Delta A_2NC_2$. Therefore $\Delta A_2B_2C_2\sim \Delta A_2MN$ and we are done.

Back to the theorem:

Let $X, Y, Z, H_2$ be the midpoints of $AA_1, BB_1, CC_1, HH_1$. It’s easy to show that $\Delta A_1H_1B_1\sim \Delta AHB, \Delta B_1H_1C_1\sim \Delta BHC, \Delta C_1H_1A_1\sim \Delta CHA$. Applying the lemma above we get $\Delta XYZ\sim \Delta ABC\sim \Delta A_1B_1C_1$ and $H_2$ is the orthocenter of $\Delta XYZ$.
Let $L$ be the Miquel point of the completed quadrilateral $ABCA'B'C'$.
The result $X, Y, Z, O, L$ are concyclic is well-known so I leave it to the readers.
Note that $A', B', C'$ is the reflections of $L$ wrt $YZ, ZX, XY$, respectively then $d$ is the Steiner’s line of $L$ wrt $\Delta XYZ$. This means $d$ passes through the orthocenter of triangle $XYZ$ or $d$ passes through the midpoint of $HH_1$.
We are done.

Another proof (by Jean-Louis Ayme):

Let $X$ be the intersection of $AA_1$ and $CC_1$. Note that $\angle BCX= \angle A'B'C_1=\angle C'AA_1$ then $X\in (ABC)$.

Let $M, N$ be the projections of $X$ onto $BC, AB$, respectively.

We have $\angle BMN=\angle BXN=\angle BB_1C'=\angle BA'C'$ therefore $MN//d$ or the Simson’s line and the Steiner’s line of $X$ wrt $\Delta ABC$ is parallel to $d$.

Similarly, the Steiner’s line of $X$ wrt $\Delta A_1B_1C_1$ is parallel to $d$.

Let $P, Q$ be the reflections of $X$ wrt $AB, A_1B_1$, respectively. It’s easy to show that $C'$ is the midpoint of $PQ$. Note that $d$ is the midline of the trapezium $PHH_1Q$ then $d$ bisects $HH_1$. We are done.