If AB=2R1 and CD=2R2 and BC=d then:TU^2 =(R2-R1)^2 + (R2+R1+d)^2 = 2R2^2+2R1^2+2d(R1+R2)+d^2 whie AC*Bd=(2R1+d)(2R2+d)=4R1R2+2d(R1+R2)+d^2. Now if TU^2=AC*BD then 2R1^2+2R2^2=4R1R2 or (R2 -R1)^2=0 or if R1=R2. So, in my opinion, the hypothesis is true only in case of equal circles. Would that not be so? Ajit
Yes u r perfectly correct! A correct diagram showns that: TU^2 = (R2+R1+d)^2-(R2-R1)^2=4R1R2+2d(R1+R2)+d^2 = AC*BD. So no problem here as I thought earlier. Ajit
If AB=2R1 and CD=2R2 and BC=d then:TU^2 =(R2-R1)^2 + (R2+R1+d)^2 = 2R2^2+2R1^2+2d(R1+R2)+d^2
ReplyDeletewhie AC*Bd=(2R1+d)(2R2+d)=4R1R2+2d(R1+R2)+d^2. Now if TU^2=AC*BD then 2R1^2+2R2^2=4R1R2 or (R2 -R1)^2=0 or if R1=R2. So, in my opinion, the hypothesis is true only in case of equal circles. Would that not be so?
Ajit
TU^2 =(R2-R1)^2 + (R2+R1+d)^2 must be:
ReplyDeleteTU^2 = - (R2-R1)^2 + (R2+R1+d)^2
rest will follow...
-binary
Yes u r perfectly correct!
ReplyDeleteA correct diagram showns that:
TU^2 = (R2+R1+d)^2-(R2-R1)^2=4R1R2+2d(R1+R2)+d^2 = AC*BD.
So no problem here as I thought earlier.
Ajit