## Wednesday, November 12, 2014

### Geometry Problem 1059: Triangle, Circumcircle, Orthocenter, Altitude, Perpendicular, Reflection of a Point over a Line, Collinear Points, Simson Line, Congruence

Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it. 1. http://s25.postimg.org/a4t0mmxmn/pro_1059.png

Let B’ and C’ are the reflection of H over AC and AB. B’ and C’ will be on circle O.
Let HD2 and HD3 cut AC and AB at E and F
HDD2B’ and HDD3C’ are isosceles trapezoids and Triangles HB’E and HFC’ are isosceles.
∠ (B’HE)+ ∠ (C’HF) = ∠ (HB’E) + ∠ (HC’F)= angle A => ∠ (C’HB’) supplement to ( ∠ (B’HE)+ ∠ (C’HF))
So E, H and F are collinear => D2, H, D3 are collinear
M, I and S is located on linear Simpson line so D1, D2 D3 and H are collinear and DF/DH= Di/ D1= 1/2
Obvious D2D3=2. SM

2. Firstly, since DM⊥AB, DI⊥BC, DS⊥AC,
thus MIS is the Simpson's line of point D.

Now since DM=D₃M, DI=D₁I, DS=D₂S,
thus D₃D₁D₂ // MIS, with D₂D₃=2×SM.

Join DH, and DH cuts MIS at F.
We want to show that H lies on D₃D₁D₂.
If so, then F is automatically the mid-point of DH and we are done.

Let BH cuts AC at J, and cuts the circumcircle at K.
Join DK. Let DK cuts AC at L, and cuts MS at N.

Since DISC concyclic, thus ∠ISD=∠ICD=∠BCD=∠BKD=∠KDS,
so ∠NDS=∠NSD, and so NS=ND.
Which means N is the circumcenter of ΔLDS, thus DN=NL.
Therefore L lies on the line D₃D₁D₂.

Now since HJ=JK, thus ∠HLJ=∠KLJ=∠DLS=∠NSL.
So HL//NS which is the Simpson's line MIS.
Hence, H lies on D₃D₁D₂ and then F is the mid-point of DH.