## Tuesday, December 17, 2013

### Geometry Problem 943: Triangle, Incircle, Inscribed Circle, Tangency Points, Parallel, Metric Relations

Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to enlarge the problem 943.

1. Since AD=AF and DF//BG, so BD=FG.
Since CD=CF and EF//BH, so BE=HF.
Since BD=BE, thus HF=FG. i.e. F is mid-point of HF.

Since JF//BH, so J is the mid-point of BG.
Since KF//BG, so K is the mid-point of BH.
Consequently, JK//GH.

It follows that KHFJ is a parallelogram.
Since F is the mid-point of GH, so L is the mid-point of JK.
Hence, LM = 1/2 JF = 1/4 BH = 5/2

2. Observe that, AF=AD=s-a, CF=CE=s-c, BE=BD=s-b [where ‘s’ is the semi perimeter.
Let, AH=p and CG=q, then, FH=s-a+p and FG=s-c-q
Now, of ΔCHB, EF||BH, so, CE/EB=CF/FH=>(s-c)/(s-b)=(s-c)/(s-a+p)=>FH=s-b
And, of ΔAGB, DF||BG, so AD/DB=AF/FG=>(s-a)/(s-b)=(s-a)/(s-c-q)=>FG=s-b
So, of ΔBHG, HF=FG=1/2 HG=s-b
So, F is the midpoint of HG and KF||BG and JF||BH,
So, ΔKFJ is the median triangle of ΔBHG, so, KH=1/2 BH=5
Now, BJ||KF and BK||JF, so ▟BKFJ is parallelogram, and BF and KJ are it’s diagonals, so, KL=LJ
Similarly, ▟KHFJ is parallelogram and KF and HJ are it’s diagonals, so, HM=MJ
Now, of ΔJKH, L and M are the midpoints of the side JK and JH respectively.
So, LM||KH and LM=1/2 KH=5/2 (ANS)