Tuesday, July 28, 2009

Problem 328. Triangle, Incircle, Tangency Points, Parallel, Midpoint

Proposed Problem
Click the figure below to see the complete problem 328 about Triangle, Incircle, Tangency Points, Parallel, Midpoint.

 Problem 328. Triangle, Incircle, Tangency Points, Parallel, Midpoint.
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Complete Problem 328
Level: High School, SAT Prep, College geometry


  1. Join B to I and extend it to meet AH. We’ve DE perpendicular to BI and since DE is parallel to AH, the latter (AH) is perpendicular to BI and by Honsberger Concurrency, FE, BI and AH are concurrent in H.
    Similarly C, I and G are collinear and AG is perpendicular to CG. Further, quad. AGIH is concyclic and thus angle AGH = angle AIH = A/2 + B/2 = 90 – C/2. But angle AGH = angle FHN since AG & FE are parallel.Thus, angle FHN = 90 – C/2 and this makes MN parallel to BC since angle FEC also (90 –C/2).
    From rt. Triangle AGC we’ve AG = bsin(C/2) hence HE = bsin(C/2) while FE = 2(s-c)sin(C/2) since CF = s-c. Therefore, FH =2(s-c)sin(C/2)-bsin(C/2)
    = sin(C/2)[2s -2c –b] = a+b+c-2c–b]=(a-c)sin(C/2)
    Further in Tr. FHN, FH/sin(C) = FN/(sin(90-C/2)
    Or(a-c)sin(C/2)=FN*sin(C)/sin(90-C/2)= 2*FN*sin(C/2) Or FN = (a-c)/2. Hence AN = AF + FN = s –a + (a-c)/2 =(a+b+c)/2–a+(a-c)/2 = b/2
    Or N is the midpoint of AC and since MN is parallel to BC, M is the midpoint of AB. QED.
    Ajit: ajitathle@gmail.com

  2. mAHF=mGEH=mADF => A,D,H,F concyclic
    similarly A,G,D,H concyclic
    thus G,D,H,F concyclic
    mGHF=mGDF=mBEF => GH//BC
    AM/MB=AG/HE=1 <=> M midpoint of AB
    similarly N midpoint of AC