## Monday, March 7, 2011

### Problem 592: Triangle, Incenter, Incircle, Tangency Point, Midpoints, Concurrent Lines, Congruence

Geometry Problem
Click the figure below to see the complete problem 592.

Zoom at: Geometry Problem 592

1. Let MO extended meet the incircle at F.
By Problem 591, BF = r =OD
So BFDO is a parallelogram and its diagonals
BD, FO bisect each other at E, say
But the middle line GH bisects BD.
Hence BD, GH, MO (extended) concur at E.

2. Refer to lines 1 and 2 of your solution “Let MO extended meet the incircle at F.
By Problem 591, BF = r =OD ”. Note that per problem 591, MO extended to meet the altitude from B, not the incircle . It is not clear to me. Please show more detail and explanation. .
Refer to line 3 of your solution “ So BFDO is a parallelogram “ . I think that it is incorrect. BF do not parallel to DO . Please show more detail.

Peter Tran

3. To Pravin
Why altitude from B passes at F (according to P591)

4. Thank you Peter Tran & c.t.e.o !

Typo error to be corrected thus:
"Let MO extended meet BE
(the altitude from A) at F"

Now BF = r = OD and BF // OD and
BFDO is a parallegram etc ...