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Level: Mathematics Education, High School, Honors Geometry, College.
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Saturday, February 29, 2020
Geometry Problem 1455: Nagel Point, Excircles, Incircle, Congruent Segments
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Using Van Aubel's theorem
ReplyDeleteWe 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
https://photos.app.goo.gl/vEqkL8Da7ogdieyr5
ReplyDeleteLet 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