Home > Shapes > Geometry
| Perpendicular Incenters (Posted on 2018-05-06) |
 |
Let a, b, and c be the lengths of sides BC, CA, and AB respectively
of ΔABC. Let D be a point on side BC. Let I and J be the incenters
of ΔABD and ΔACD respectively.
If the line segment IJ ⊥ AD, then what is the value of the ratio
|BD| / |CD| in terms of a, b, and c?
|
|
Submitted by Bractals
|
|
No Rating
|
|
|
Solution:
|
(Hide)
|
Let E and F be the points on AD where the incircles of ΔABD and
ΔACD kiss AD. Clearly IJ ⊥ AD when E and F coincide.
|DE| = |DF|
|BD| + |AD| - |AB| |CD| + |AD| - |AC|
---------------------- = ----------------------
2
2
|BD| = ( |AB| + |BC| - |AC| ) / 2 = (c + a - b) / 2.
Note: D is the point on BC where the incircle of ΔABC kisses BC.
|CD| = (a + b - c) / 2
|BD| / |CD| = ( c + a - b ) / ( a + b - c )
QED
|
Comments: (
You must be logged in to post comments.)
There are no comments yet.
|
|
Please log in:
Forums (1)
Newest Problems
Random Problem
FAQ |
About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
|
