## Tuesday, April 7, 2009

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

Proposed Problem

See complete Problem 24 at:
gogeometry.com/problem/p024_right_triangle_incircles.htm

Level: High School, SAT Prep, College geometry

1. 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))
Likewise, 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

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

3. Reference my proof for Problem 23

r1 = 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

5. Let the point M be the center of the circle with radius r1.
Let the point N be the center of the circle with radius r2.
Let the point O be the center of the circle with radius r.

Then draw a line from M perpendicular onto BD. The line from M meets BD at the point P.

Also draw a line from the point N perpendicular onto the radius r. This line meets

We have to use the results from problem 23 BD=r1+r2+r and the result from problem 28 DE =r2-r1 .

PB = BD-r1 = r+r2
QN = r2-DE = r1

And

OQ = r-r2
MP = r1

Triangles BPM and NQO are similar we have :

MP/BP = OQ/NQ
r1/(r+r2) = (r-r2)/r1
r1^2 = (r+r2)(r-r2)
r1^2 = r^2-r2^2

r1^2+r2^2=r^2