Proposed Problem

See complete Problem 24 at:

gogeometry.com/problem/p024_right_triangle_incircles.htm

Level: High School, SAT Prep, College geometry

## Tuesday, April 7, 2009

### Problem 24: Right triangle, Altitude, Incircles

Labels:
altitude,
incircle,
inradius,
Pythagoras,
right triangle,
similarity

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With the usual nomenclature, we've BD=ac/b and AD = c^2/b. Thus, r1 = (ca/b * c^2/b)/(ac/b+c^2/b+c) = ac^2/(b(a+b+c))

ReplyDeleteLikewise, r2 = ca^2/(b(a+b+c))

Therefore, r1^2 + r2^2 =

a^2c^4/(b(a+b+c))^2 + c^2a^4/(b(a+b+c))^2

= a^2c^2(a^2+c^2)/(b(a+b+c))^2

= a^2c^2b^2/(b(a+b+c))^2

= [ac/(a+b+c)]^2 = r^2

QED

Ajit: ajitathle@gmail.com

Triangle ABD similar to tri. ABC and we have r1/r = AB/AC

ReplyDeleteTriangle BDC similar to tri. ABC and we have r2/r = BC/AC

ABC is a right triangle so we have AC^2=AB^2+BC^2

And r^2=r1^2+r2^2

Peter Tran

Reference my proof for Problem 23

ReplyDeleter1 = rc/b and r2 = rc /b

So r1^2 + r2^2 = r^2(a^2 + c^2)/b^2 = r^2 from Pythagoras

Sumith Peiris

Moratuwa

Sri Lanka

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ReplyDelete