All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 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.)
 Subject Author Date
There are no comments yet.

 Search: Search body:
Forums (0)