## Monday, March 4, 2019

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

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below. 1. https://photos.app.goo.gl/YvmQY9G5cGXqPdkX9

We 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

1. Minor typo error BI=SQRT(2)/3
Peter tran

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

3. I am not getting thought proces similarity

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

5. https://photos.app.goo.gl/PxGXK7bKbXqV3Zi6A

Per 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

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

From 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