Let 
be the circumradius of triangle 
and let 
be the circumradius of triangle 
Compute 
![[asy] unitsize(1 cm);pair A, B, C, D;A = (-2,0);B = (0,5);C = (2,0);D = (-1,0);draw(A--B--C--cycle);draw(B--D);label(](/api/ssl-img-proxy?src=https%3A%2F%2Flatex.artofproblemsolving.com%2F3%2F9%2Fe%2F39e843ba85c6f058e70b5340ef245dbe05a038c6.png)
Help
+130552 I don't know the most direct method of this.....but
The circumradius of a triangle is given by : (product of the sides) / [ 4 * area ]
Since the two interior triangles are under the same height, they are each other as to their bases
So area of BDC = 3 area of BDA
So, since BD is common to both, the product of the sides of BDC = 3 times the product of the sides of BDA
So
R1 / R2 = [ product of the sides of BDA ] / [ 4 *area of BDA ]
________________________________________________ =
[ 3 * product of the sides of BDA ] / [ 4 * (3 *area of BDA ) ]
[ product of the sides of BDA ] * [ 4*( 3 * area of BDA) ]
______________________________________________ = 12 / 12 = 1
[ 3 * product of the sides of BDA] * [ 4 *area of BDA ]
