Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

## Monday, March 4, 2019

### Geometry Problem 1421: Right Triangle, Incircle, Excircle, Tangent Lines, Measurement

Labels:
excircle,
geometry problem,
incircle,
measurement,
right triangle,
tangent

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https://photos.app.goo.gl/YvmQY9G5cGXqPdkX9

ReplyDeleteWe have s= ½(3+4+5)=6

BD=FC=s-AC= 1=> BI= sqrt(2)

Since GF//IC=> BG/BI= BF/FC =2/3

So BG= 2.sqrt(2)/3 and BI= sqrt(2)/2

Minor typo error BI=SQRT(2)/3

DeletePeter tran

Sir may I know the book you refer for geometry the way solved the question is outstanding

DeleteI am not getting thought proces similarity

Deletethe similarity in triBGF AND triBIC u applied but how can we say GF is parallel to IC

Deletehttps://photos.app.goo.gl/PxGXK7bKbXqV3Zi6A

DeletePer the result of problem 1407, we have D, F, M are collinear ( see sketch above)

and IC ⊥ CE and IC ⊥ FM so IC // GF

Peter Tran

Let EF & BGI extended meet at X. Let the tangency point of incircle on BC be Y

ReplyDeleteFrom previous problems, BDIY is a square of side b-c = 1 and so FY = 1 and BX = 2.sqrt2

BD/FX = BG/GX, 1/2 = BG/(2.sqrt2 - BG) since BX = 2.sqrt2

Hence BG = 2.sqrt2/3 and since BI = sqrt2,

IG = sqrt2/3

Sumith Peiris

Moratuwa

Sri Lanka