Proposed Problem

See also:Complete Problem 28, Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Saturday, April 11, 2009

### Problem 28. Right Triangle, altitude, incircles, inradius

Labels:
altitude,
common tangent,
incircle,
inradius,
right triangle,
tangency point

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Let incircle of Tr. ABC meet AB in H. Thus, AE = AH = AB - BH = c - r = c - (c+a-b)/2 = (c-a+b)/2.

ReplyDeleteFrom Tr ABD, AD = c^2/b. Hence, DE = AE - AD = (c-a+b)/2 -c^2/b = (bc -ab+b^2-2c^2)/2 ------(1)

Triangles ABD & CBD give 2r2=CD+BD -a and 2r1= AD+BD -c whence 2r2 -2r1 = CD - AD + c - a = a^2/b - c^2/b + c -a or r2-r1=(bc-ab+a^2-c^2)/2b = (bc-ab+b^2-2c^2)/2---(2) since b^2=a^2+c^2. Equations (1) & (2) give us r2 -r1 = DE whereas it is appaent from the diagram that FG =r2 - r1. Hence, DE = FG = r2 - r1

Vihaan: vihaanup@gmail.com

AB=a , BC =b , AD= n , DC= m and BD = h

ReplyDeleteFor the radius of the circles we have the following equations :

2r=a+b-m-n

2r1 = n+h-a

2r2 =m+h-b

DC= m and EC= b-r so we get :

DE =DC-EC=m-(b-r)

using the formula for 2r :

2DE=2m-2b+2r

2DE=2m-2b+a+b-n-m

2DE=m-b+a-n

2DE=m+h-b+a-h-n

2DE=m+h-b-(n+h-a)

m+h-b=2r2 and n+h-a =2r1 therefore we get :

2DE=2r2-2r1

DE=r2-r1