Problem:
Find the area of rhombus ABCD given that the radii of the circles circumscribed around triangles ABD and ACD are 12.5 and 25 , respectively.
Solution:
Let O be the point of intersection of diagonals AC and BD, and E the point of intersection of AC and the circumcircle of △ABD. Extend DB to meet the circumcircle of △ACD at F. From the Power-of-a-Point Theorem, we have
AO⋅OE=BO⋅OD and DO⋅OF=AO⋅OC
Let AC=2m and BD=2n. Because AE is a diameter of the circumcircle of △ABD, and DF is a diameter of the circumcircle of △ACD, the above equalities can be rewritten as
m(25−m)=n2 and n(50−n)=m2
or
25m=m2+n2 and 50n=m2+n2
Therefore m=2n. It follows that 50n=5n2, so n=10 and m=20. Thus [ABCD]=(1/2)AC⋅BD=2mn=400.
OR
Let R1 and R2 be the circumradii of triangles ABD and ACD, respectively. Because BO is the altitude to the hypotenuse of right △ABE,AB2=AO⋅AE. Similarly, in right △DAF,AB2=DA2=DO⋅DF, so AO⋅AE=DO⋅DF. Thus
DOAO=AEDF=R1R2=2
Also, from right △ADE,2=DOAO=OEDO. Then
25=2R1=AE=AO+OE=2⋅DO+21DO=25DO
and DO=10,AO=20, so [ABCD]=400.
OR
Let s be the length of a side of the rhombus, and let α=∠BAC. Then AO=scosα, and BO=ssinα, so [ABCD]=4[ABO]=2s2sinαcosα=s2sin2α. Apply the Extended Law of Sines (In any △ABC with AB=c,BC=a, CA=b, and circumradius R,sinAa=sinBb=sinCc=2R) in △ABD and △ACD to obtain s=2R1sin(90∘−α)=sin2R1cosα, and s=2R2sinα. Thus tanα=cosαsinα=R2R1=21. Also, s2=4R1R2cosαsinα=2R1R2sin2α. But sin2α=2⋅51⋅52=54, from which [ABCD]=2R1R2sin22α=2⋅225⋅25⋅2516=400 .
OR
Let AB=s,AO=m, and BO=n, and use the fact that the product of the lengths of the sides of a triangle is four times the product of its area and its circumradius to obtain 4[ABD]R1=s⋅s⋅2n and 4[ACD]R2=s⋅s⋅2m. Since [ABD]=[ACD], conclude that 21=R2R1=mn, and proceed as above.
