Let the angle = Φ, so we've b^2+a^2-2abcos(Φ)=c^2+b^2-2bccos(Φ) since the line segments opposite must also be equal. This gives us, cos(Φ)=(c^2-a^2)/(2b(c-a)) =(c+a)/(2b). Likewise, cos(Φ)=(d^2-b^2)/(2c(d-b)) =(b+d)/(2c) or (a+c)/(2b)=(b+d)/(2c) or (a+c)/(b+d) =b/c Ajit
Let the angle = Φ, so we've b^2+a^2-2abcos(Φ)=c^2+b^2-2bccos(Φ) since the line segments opposite must also be equal. This gives us, cos(Φ)=(c^2-a^2)/(2b(c-a)) =(c+a)/(2b).
ReplyDeleteLikewise, cos(Φ)=(d^2-b^2)/(2c(d-b)) =(b+d)/(2c) or (a+c)/(2b)=(b+d)/(2c) or (a+c)/(b+d) =b/c
Ajit
Look at fox 298 newzad's solution,same way, just connect the blue points on circle
ReplyDeleteBleaug says:
ReplyDeletefrom #300: cos α = (a+c)/2b = (b+d)/2c which is equivalent to (a+c)/(b+d) = b/c