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由 $\left(\sqrt{b^2+c^2}+b-c\right)\left(\sqrt{a^2+c^2}+a-c\right)=2ab$ 可得$$\dfrac{\sqrt{b^2+c^2}+b-c}{b}=\dfrac{2a}{\sqrt{a^2+c^2}+a-c}=\dfrac{\sqrt{a^2+c^2}-a+c}{c}.$$所以$$c\left(\sqrt{b^2+c^2}+b-c\right)=b\left(\sqrt{a^2+c^2}-a+c\right),$$整理即$$ab-c^2=b\sqrt{a^2+c^2}-c\sqrt{b^2+c^2}.$$两边平方整理得$$\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}=ac+bc,$$两边平方得$$c^4-2abc^2+a^2b^2=0,$$即$$\left(c^2-ab\right)^2=0,$$所以$$c^2=ab.$$由 $\left(\sqrt{b^2+c^2}+b-c\right)\left(\sqrt{a^2+c^2}+a-c\right)=2ab$ 可得$$\dfrac{\sqrt{b^2+c^2}+b-c}{b}\cdot \dfrac{\sqrt{a^2+c^2}+a-c}{a}=2,$$即$$\left(\sqrt{1+\left(\dfrac cb\right)^2}+1-\dfrac cb\right)\left(\sqrt{1+\left(\dfrac ca\right)^2}+1-\dfrac ca\right)=2$$令$$f(x)=\sqrt{1+x^2}+1-x,$$则$$f\left(\dfrac cb\right)\cdot f\left(\dfrac ca\right)=2.$$因为$$\begin{split}\dfrac{2}{f(x)}&=\dfrac 2{\sqrt{1+x^2}+1-x}\\&=\dfrac{\sqrt{1+x^2}-1+x}{x}\\&=\sqrt{1+\dfrac 1{x^2}}-\dfrac 1x+1\\&=f\left(\dfrac 1x\right),\end{split}$$所以$$f\left(\dfrac cb\right)=\dfrac 2{f\left(\dfrac ca\right)}=f\left(\dfrac ac\right).$$而函数$$f(x)=\sqrt{1+x^2}-x+1=\dfrac 1{\sqrt{1+x^2}+x}+1,$$所以 $f(x)$ 在 $(0,+\infty)$ 上单调递减，因此$$\dfrac cb=\dfrac ac,$$即$$c^2=ab.$$