Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to enlarge the problem 943.

## Tuesday, December 17, 2013

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

Labels:
incircle,
inscribed,
metric relations,
parallel,
tangency point,
triangle

Subscribe to:
Post Comments (Atom)

Since AD=AF and DF//BG, so BD=FG.

ReplyDeleteSince 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

Observe that, AF=AD=s-a, CF=CE=s-c, BE=BD=s-b [where ‘s’ is the semi perimeter.

ReplyDeleteLet, 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)