Proposed Problem

See complete Problem 280 at:

http://gogeometry.com/problem/p020_right_triangle_incircle.htm

Level: High School, SAT Prep, College geometry

## Saturday, April 4, 2009

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

Labels:
altitude,
inradius,
perpendicular,
right triangle

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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.

ReplyDeleteUsing 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

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

ReplyDeleteAll triangles AED, EDF and DFC are similar and we have

ReplyDeleteAE/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

Note that △ADE is similar to both △DCF and △EFD (AAA)

ReplyDeleteSimilar 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.