The equation is homogenous, so let the hypotenuse be $1$ with $\angle ABC=\theta$. We get $$AD\cdot BC=\sin\theta\cos\theta,$$$$AB\cdot CD=\cos\theta\sin^2\theta,$$$$BD\cdot AN=\cos^2\theta\frac{\sin\theta\cos\theta\sin\theta}{\sin\theta+\sin^2\theta}.$$Now we have $$\sin^2\theta\cos\theta+\frac{\sin\theta\cos^3\theta}{1+\sin\theta}=\frac{\sin^2\theta\cos\theta+\sin\theta\cos\theta\left(\sin^2\theta+\cos^2\theta\right)}{1+\sin\theta}=\sin\theta\cos\theta.$$