Proposed Problem

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

See more:

Complete Problem 328

Level: High School, SAT Prep, College geometry

## Tuesday, July 28, 2009

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

Subscribe to:
Post Comments (Atom)

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.

ReplyDeleteSimilarly 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

mAHF=mGEH=mADF => A,D,H,F concyclic

ReplyDeletesimilarly 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