## Saturday, April 4, 2009

### Problem 20: Right Triangle, Altitude, Inradii, Perpendicular

Proposed Problem See complete Problem 280 at:
http://gogeometry.com/problem/p020_right_triangle_incircle.htm

Level: High School, SAT Prep, College geometry

1. Using similarity between various rt. triangles that are there in the given diagram, it can be shown that: AD=AB^2/AC,CD=BC^2/AC, AE=AB^3/AC^2, DE=AB^2*BC/AC^2,DF=(AB*BC/AC)^2 and EF=BC*AB/AC.
Using these relations and the fact that inradius = Tr.Area/Semisum of sides, we can show that:
a = BC*AB^3/(AB+BC+CA)AC^2
Likewise, b = AB*BC^3/(AB+BC+CA)AC^2
while x =(BC*AB/AC)^2*(1/(AB+BC+CA)
From these relations we can conclude that x^2=a*b
QED
Ajit: ajitathle@gmail.com

2. ahmetelmas-geo-geo-antonio.blogspot.com/

3. All triangles AED, EDF and DFC are similar and we have
AE/ED=a/x
AE/DF=AE/EB=a/b
In right triangle ADB we have ED^2=AE.EB so x^2=a.b

Peter Tran

4. Note that △ADE is similar to both △DCF and △EFD (AAA)
Similar triangles implies similar incircles.
Let DE=c,DF=g,CF=d.
From above, we can deduce:
1. a/x=c/g
2. x/b=g/d
3. c/g=g/d (since DE/DF=DF/CF)
Substitute equation 1 and 2 into equation 3,
a/x=x/b
ab=x^2
x=SQRT(ab)
proved.