Problem

Let ABC be a triangle with
I
A
,
I
B
, and
I
C
as excenters.
Prove that excentral triangle
I
A
I
B
I
C
has orthocenter
I
H
and that triangle ABC is its orthic triangle ( pedal triangle centered on orthocenter ).

Geometric Scene Construction

Though we can use GeometricScene to generate arbitrary triangle, we can specify some coordinates for our demo triangle to make the diagram with visually pleasant ratio.
This code is rather simple with repetitive TriangleCenter calls.
gs=GeometricScene[{a{0,0},b{4,0},c{3.1,3.5},ia,ib,ic},{​​ia==TriangleCenter[{a,b,c},{"Excenter",a}],​​ib==TriangleCenter[{a,b,c},{"Excenter",b}],​​ic==TriangleCenter[{a,b,c},{"Excenter",c}],​​TriangleConstruct[{ia,ib,ic},"Orthocenter"]​​}]//RandomInstance;​​Framed[gs]
Out[]=

Proof by Angle Chasing

◼
  • Colinear property. WLOG, you only need
    I
    B
    C
    I
    A
    sit on the same line. Notice that
    ∠I
    B
    CA=
    ∠I
    A
    CB
    . We can quickly check
    • ∠I
    B
    CA
    is
    1
    2
    (π-∠BCA)
    PlanarAngle[{ib,c,a}]/.gs["Points"]
    Out[]=
    1.08252
    and
    ∠I
    A
    CB
    is
    1
    2
    (π-∠BCA)
    PlanarAngle[{ia,c,b}]/.gs["Points"]
    Out[]=
    1.08252
    which is very clear
    ∠I
    A
    CI
    B
    =
    ∠I
    B
    CA+∠BCA+
    ∠I
    A
    CB=π
    . Done.
    ◼
  • Perpendicular property. WLOG, we only need line
    CI
    c
    is perpendicular to
    I
    A
    I
    B
    .
    This is simple because we know
    I
    C
    is excenter, it implies that
    CI
    c
    is angle bisector of
    ∠BCA
    .
    ∠I
    B
    CI
    C
    =
    ∠I
    B
    CA+
    ∠I
    C
    CA=
    ∠I
    C
    CB+
    ∠I
    A
    CB=
    ∠I
    B
    CI
    A
    .
    We also know
    ∠I
    B
    CI
    C
    +
    ∠I
    C
    CI
    A
    =
    ∠I
    A
    CI
    B
    =π
    . Thus,
    ∠I
    B
    CI
    C
    =π/2
    . Done.
    • PlanarAngle[{ib,c,a}]+PlanarAngle[{ic,c,a}]/.gs["Points"]
    Out[]=
    1.5708
    %/Degree
    Out[]=
    90.
    Therefore we have
    BI
    B
    ⊥
    I
    C
    I
    A
    and
    AI
    A
    ⊥
    I
    C
    I
    B
    . This implies triangle ABC is the orthic triangle of triangle
    I
    A
    I
    B
    I
    C
    .
    ◼
  • Additionally, we have some similarity property. WLOG, we only have to prove
    CAI
    B
    ∼
    BCI
    A
    . ​
    ​We need one more angle equation besides
    ∠I
    B
    CA=
    ∠I
    A
    CB
    . Notice that
    • ∠I
    A
    BC=
    1
    2
    (π-∠ABC)and
    ∠AI
    B
    C=
    1
    2
    (π-
    ∠I
    B
    AC-
    ∠I
    B
    CA)=
    1
    2
    π-
    1
    2
    (π-∠BAC)-
    1
    2
    (π-∠BCA)​​=
    1
    2
    (∠BAC+∠ACB)=
    1
    2
    (π-∠ABC)=
    ∠I
    A
    BC
    Thus
    CAI
    B
    ∼
    BCI
    A
    . We can prove other similar triangle pairs in the same way. ​
    ​Use FindGeometricConjectures to list them
    FindGeometricConjectures[gs,_[__,"Similar"]]["Conclusions"]
    Out[]=
    {GeometricAssertion[{Triangle[{ia,b,c}],Triangle[{a,ib,c}],Triangle[{a,b,ic}],Triangle[{ia,ib,ic}]},Similar]}

    Review

    ◼
  • Use excenter’s definition and angle bisectors
    • ◼
  • Arithmetic with neighboring angles to find perpendicular segments
    • ◼
  • Find similar triangles with angular addition and interior angles in a triangle sum up to π

    • Other Geometry Problems

    ◼
  • Summable Areas
    • ◼
  • Circumcenter, Midpoint and Pi