Geometry Problem

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## Monday, March 7, 2011

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

Labels:
concurrent,
congruence,
incenter,
incircle,
midpoint,
tangency point,
triangle

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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 ...