Question

One can argue this question can in geometry, but this property of the Arbelo is elementary and a good foundation for intuitive and observational proofs, so show that that the length of the lower boundary of the arbelos equals the length upper boundary?

Answer 1

Calling `hat(AB)` the semicircumference length with radius `r`, `hat(AC)` the semicircumference length of radius `r_1` and `hat(CB)` the semicircumference length with radius `r_2`

We know that

`hat(AB) = lambda r`, `hat(AC) = lambda r_1` and `hat(CB)= lambda r_2` then

`hat(AB)/r=hat(AC)/r_1=hat(CB)/r_2` but

`hat(AB)/r = (hat(AC)+hat(CB))/(r_1+r_2) = (hat(AC)+hat(CB))/r`

because if

`n_1/n_2=m_1/m_2 = lambda` then

`lambda =( n_1pmm_1)/(n_2pmm_2) = (lambda n_2pm lambda m_2)/(n_2pmm_2) = lambda`

so

`hat(AB)= hat(AC)+hat(CB)`