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2014-11-17T17:41:24+05:30

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Look at the diagram.

ID, IE, IF are radii of incircle and perpendiculars on to the sides from incenter I. 

Triangles AIE, AIF are similar as AI is common, IE=IF, and angles IEA = IFA = 90 deg.  Also, AE=AF as tangents from same point A to incircle.
Similarly, triangles BIF, BID are similar.  Also,  CID, and CIE are similar. 

Hence, angle AIF=AIE= z.
   In the quadrilateral AFIE,      angle A = 360-(90+90+FIE) = 180  - 2 z
Similarly from quadrilateral BFID, we get         angle B = 180 - 2 x
From quadrilateral CDIE,  we get    angle C = 180 - 2 y
 
   We are given that BIC : CIA : AIB  = 5 : 6 : 7
 
           x+y :  y+z  :  z+x  = 5 : 6 : 7

            (x+y)6 = (y+z)5    =>  y + 6x = 5z 
            (x+y)7 = (z+x) 5   =>  2x + 7y = 5z 
 
               solving these simultaneous equations, we get 6 y = 4 x, or 3 y = 2 x
                      and 10 y = 5z,   z = 2y 

angle A = 180 - 2 z = 180 - 4y,       angle B = 180 - 2 x = 180 - 3y
angle C = 180 - 2 y

Their sum = A+B+C = 180 =>   9 y = 360,  y = 40 deg.

angle A = 20 deg,   angle B = 60 deg     angle C = 100 deg.

       A : B : C  = 1 : 3 : 5

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