Geometry Problem
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Monday, March 7, 2011
Problem 592: Triangle, Incenter, Incircle, Tangency Point, Midpoints, Concurrent Lines, Congruence
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Let MO extended meet the incircle at F.
ReplyDeleteBy 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.
Refer to lines 1 and 2 of your solution “Let MO extended meet the incircle at F.
ReplyDeleteBy 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
To Pravin
ReplyDeleteWhy altitude from B passes at F (according to P591)
Thank you Peter Tran & c.t.e.o !
ReplyDeleteTypo 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 ...