Let

sides of $\displaystyle A:x,y,z$

sides of $\displaystyle B: \alpha, \gamma, y$

sides of $\displaystyle C: \alpha, \beta, x$

sides of $\displaystyle D: \beta, \gamma, z$.

Then we have

$\displaystyle B=\frac{1}{2}\alpha \gamma, \, C=\frac{1}{2}\alpha \beta , \, D=\frac{1}{2}\beta \gamma$

and

$\displaystyle x^2=\alpha^2+\beta^2$

$\displaystyle y^2=\alpha^2+\gamma^2$

$\displaystyle z^2=\beta^2+\gamma^2.$

Using Heron's formula, we have

$\displaystyle A^2=\frac{1}{16}(x+y+z)(y+z-x)(x+y-z)(z+x-y)$

$\displaystyle =\frac{1}{16}((y+z)+x)((y+z)-x)(x+(y-z))(x-(y-z))$

$\displaystyle =\frac{1}{16}((y+z)^2-x^2)(x^2-(y-z)^2)$

$\displaystyle =\frac{1}{16}(2yz-(x^2-y^2-z^2))(2yz+(x^2-y^2-z^2))$

$\displaystyle =\frac{1}{16}(4y^2z^2-(x^2-y^2-z^2)^2)$

$\displaystyle =\frac{1}{16}\left[4(\alpha^2+\gamma^2)(\beta^2+\gamma^2)-((\alpha^2+\beta^2)-(\alpha^2+\gamma^2)-(\beta^2+\gamma^2))^2\right]$

$\displaystyle =\frac{1}{16}\left[4(\alpha^2\beta^2+\beta^2\gamma^2+\alpha^2\gamma^2 +\gamma^4)-4\gamma^4\right]$

$\displaystyle =\frac{1}{4}\alpha^2\beta^2+\frac{1}{4}\beta^2\gam ma^2+\frac{1}{4}\alpha^2\gamma^2=C^2+D^2+B^2.$