Let ABCD be a cyclic quadrilateral. Let r_A, r_B, r_C, and r_D be the inradii of triangles BCD, CDA, DAB, and ABC respectively.

Prove that r_A + r_C = r_B + r_D.

Hi!

To prove to problem i use, once again, the relation :

In a triangle   cos(A)+cos(B)+cos(C) = 1 + r/R (1)

See

http://perplexus.info/show.php?pid=5987&cid=39985

All the triangle involve in the problem (BCD, CDA, DAB, ABC) have the same cyclic radius (R).

Let note

<BAC = A2 ;  <CAD = A1

< CBD = B2 ; <ABD = B1

<ACB = C1 ; <ACD = C2

<BDC = D1 ; <BDA = D2

From (1) relation we find 

r(A) = R(cos(C)+cos(B2)+cos(D1)-1) ; r(B) = R(cos(D)+cos(C2)+cos(A1)-1)

r(C) = R(cos(A)+cos(B1)+cos(D2)-1) ; r(D) = R(cos(B)+cos(A2)+cos(C1)-1)

So the relation r(A)+r(C)=r(B)+r(D) became

cos(C)+cos(B2)+cos(D1)+cos(A)+cos(B1)+cos(D2) = cos(D)+cos(C2)+cos(A1)+cos(B)+cos(A2)+cos(C1)

This relation is obvious because

<A1=<B2 ; <A2=<D1 ; <B1=<C2 ; <C1=<D2 and  cos(C)+cos(A)=0 ; cos(B) + cos(D) = 0.

Hence proved!

 

Edited on April 3, 2008, 11:59 am

Posted by Chesca Ciprian on 2008-04-03 11:56:59