Saturday, February 29, 2020

Geometry Problem 1455: Nagel Point, Excircles, Incircle, Congruent Segments

Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1455: Nagel Point, Excircles, Incircle, Congruent Segments.


  1. Using Van Aubel's theorem
    We get (AN/NTa)=(ATc/TcB)+(ATb/TbC)=((s-b)/(s-a))+ ((s-c)/(s-a))
    AN/NTa= a/(s-a), adding 1 to both sides we get ATa/NTa=s/(s-a)
    Or NTa/ATa=(s-a)/s ........(1)

    If we draw a tangent to incircle which is parallel to BC, it will touch incircl at Ia. Lets assume altitude from A to BC is h. Then we get
    AIa/ATa=(h-2r)/h, replacing h by 2S/a and r by S/s, we get
    AIa/ATa= (s-a)/s .....(2)
    Using eq (1) and (2) we have AIa= NTa
    Similary BIb=NTb and CIc= NTc


    Let a, b, c are 3 sides of triangle ABC and 2p is its perimeter
    Define point J, R(A) and r per attached sketch
    From I(A) draw B’C’// BC
    Note that AB’C’ is the image of ABC in the homothetic transformation
    Since circle J is the excircle of triangle ABC => circle I is the excircle of triangle AB’C’
    So AI(A)/ AT(A)= r/R(A)= (p-a)/p…..(1)
    Apply Van Aubel II theorem in triangle ABC with point N
    we have AN/AT(A)= (p-b)/(p-a) + (p-c)/(p-a)= a/(p-a)
    From above expression we will have NT(A)/AT(A)= (p-a)/p…..(2)
    Compare (1) to (2) we have AI(A)=NT(A)
    Similarly we will have same results with other 2 vertexes