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Sum of Quad. Radii (Posted on 2008-04-02) Difficulty: 3 of 5
Let ABCD be a cyclic quadrilateral. Let rA, rB, rC, and rD be the inradii of triangles BCD, CDA, DAB, and ABC respectively.

Prove that rA + rC = rB + rD.

  Submitted by Bractals    
Rating: 2.3333 (3 votes)
Solution: (Hide)
If T denotes triangle XYZ, the problem Sum of Radii can be stated as

    rT + RT = TX + TY + TZ
For our problem, we have

     rA + RA = AB + AC + AD

     rB + RB = BC + BD + BA

     rC + RC = CD + CA + CB

     rD + RD = DA + DB + DC
Therefore,

     rA + rC = rB + rD
follows from the fact that

     RA = RB = RC = RD,

     AB = BA, BC = CB, CD = DC, DA = AD,

     AC = -CA, and BD = -DB. 

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionRadii thriller!Chesca Ciprian2008-04-03 11:56:59
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