Clarification on a step of a proof of the isoperimetric inequality

Question

In Manfredo do Carmo's Differential Geometry of Curves and Surfaces, while proving the Isoperimetric Inequality, the following construction is employed.
We consider a simple regular closed curve and any two parallel lines that do not meet the curve. We then translate them along a perpendicular(to both of them ofc) line which connects them until they meet the curve, gaining therefore two new parallel lines, L and L', tangent to it. We then consider a circumference which does not meet the curve, with radius $r$, tangent to L and L'. We further orient the axis such that the x-axis is aligned with the line perpendicular to L and L', the y-axis runs parallel to L and L', and the center of the reference frame is the center of the circumference.
Having constructed all that, we parametrize the curve by arclength, $\alpha(s)=(x(s),y(s))$ and the circumference, with the same parameter as the curve, by $\bar{\alpha}(s)=(\bar{x}(s),\bar{y}(s))=(x(s), \bar{y}(s))$. Then the isoperimetric inequality is argued with the formula for area (bounded by the curve) given by $$A=-\int_0^l yx'ds=\int_0^lxy'ds=\frac{1}{2}\int_0^l (xy'-yx')ds$$ Using this, we arrive at the isoperimetric inequality, $4\pi A\leq l^2$, where $l$ is the length of the curve. Now, my question is about when the equality holds. do Carmo argues that this only holds when the curve is the circumference itself. To do that, he takes this inequality $$A+\pi r^2=\int_0^l (xy'-\bar{y}x')ds\leq \int_0^l \sqrt{(xy'-\bar{y}x')^2}ds\leq \int_0^l \sqrt{(x^2+\bar{y}^2)(x'^2+y'^2)}ds$$ which he had previously used and states that equality must hold everywhere on the domain of the integral, to conclude $(xy'-\bar{y}x')^2=(x^2+\bar{y}^2)(x'^2+y'^2)$ from whence we have $(xx'+\bar{y}y')^2=0$. The part that I didn't quite get is what's next: he writes $$\frac{x}{y'}=\frac{\bar{y}}{x'}=\frac{\sqrt{x^2+\bar{y}^2}}{\sqrt{x'^2+y'^2}}$$ from where he goes on to prove that this enforces $x$ and $y$ to reproduce the equation of the circumference. My question is where does the second equality come from? I tried to manipulate algebraically the equality before this one but didn't get anywhere.

Answer 1

There is a sign missing in the first equality. But if $a/b=\pm c/d=\lambda$, then $a^2+c^2=\lambda^2(b^2+d^2)$. The second equality follows, with a $\pm$ factor added to it. Note also that the ratio of square roots is positive, so the $\pm r$ is more than suspicious. — You might check out the slightly different argument on p. 31 of my differential geometry text (linked in my profile).