M is the midpoint of the side BC of triangle ABC, the center of whose
incircle is denoted by O. AH is perpendicular to BC, and the line MO intersects AH at the point E. It is known that: |AB| ≠ |AC|.
Prove that the distance AE is equal to the radius of the incircle.
Let H be the foot of the altitude from
vertex A to side BC and D the foot of
the perpendicular from O to side BC.
Let a = |BC|, b = |CA|, c = |AB|,
h = |AH|, r = |OD| (the inradius),
and s = (a+b+c)/2.
Assume b < c ( if b > c, then swap the
letters B-C and b-c in the following
proof).
|BM| = a/2
|BD| = s-b = (a-b+c)/2
|BH| = (a^2 - b^2 + c^2)/(2a)
|MD| = |BD|-|BM| = (c-b)/2
|MH| = |BH|-|BM| = (c-b)(c+b)/(2a)
From the similar right triangles MDO
and MHE:
r |OD| |EH|
--------- = ------ = ------
(c-b)/2 |MD| |MH|
|EH|
= -----------------
(c-b)(c+b)/(2a)
or
r(b+c)
|EH| = --------
a
Area of triangle ABC = rs = ah/2
or
h = 2rs/a
r(b+c)
|AE| = |AH|-|EH| = h - --------
a
2rs r(b+c)
= ----- - --------
a a
r r
= --- [2s - (b+c)] = --- [a]
a a
= r
|
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Posted by Bractals
on 2010-10-14 15:43:33 |
Solution
