Geometry Problem

Click the figure below to see the complete problem 611.

## Wednesday, June 1, 2011

Subscribe to:
Post Comments (Atom)

skip to main |
skip to sidebar
## Wednesday, June 1, 2011

###
Problem 611: Altitude of a Triangle, Perpendicular, Collinear Points, Parallel

## Link List

Online Geometry theorems, problems, solutions, and related topics.

Subscribe to:
Post Comments (Atom)

Analytic Proof:

ReplyDeleteLet D = (o,o), A = (a,0), C = (c,0), B = (0,b)

The equations of AB, DEM are respectively

bx + ay = ab and ax-by = 0

Solving, we have

E = [ab²/(a² + b²) , a²b/(a² + b²)]

Similarly,

F = [b²c/ (b² + c²), bc²/ (b² + c²)]

It can be verified that

the equation of EF is

b (a + c)x – (b² - ac)y = abc

Hence

G = [0, abc/ (ac - b²)] and

so the equation of CG is

abx + (ac - b²)y =abc

Solving it with the equation

ax-by = 0 of DE

we obtain (eliminating x)

y-coord of M = b = y-coord of B

M and B are at the same height over AC

Hence BM is parallel to AC

Similarly BH is parallel to AC

Thus M, B, H are collinear

http://img840.imageshack.us/img840/7563/problem611.png

ReplyDeleteDraw circle diameter BD ( see attached sketch)

From B draw a line parallel to AC . This line cut DE and DF at M and H

AH cut BD at G and MC cut BD at G’ .

By the properties of right triangles we have BH.DC=BD^2= BM.AD

So BM/DC=BH/AD

∆MBG similar to ∆ CDG => BG’/DG’= BM/DC

∆HBG similar to ∆ADG => BG/DG= BH/AD

From above expressions we have BG’/DG’= BG/DG => G coincide to G’

M, B and H are collinear and BH // AC