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Archive for Tháng Bảy, 2011

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.


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.


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