Help with proving a 2 by 2 determinant is the area of parallelogram I have proved a large part of this by the following but get stuck at the last step.

To say $A=ad-bc$, we still need $ad>bc$. I have puzzling over this for hours. Thank you!

In general, $A \geq 0$ is area and $A^2= (ad-bc)^2$, then $A=\vert ad-bc \vert$, so in fact area of a parallelogram is absolute value of the determinant. 
In your case, where $(c,d)$ is between 0 and 180 degrees ccw of $(a,b)$, call this angle $\theta \in $[0, 180].  Note $(d,-c)$ is a rotation of $(c,d)$ 90 degrees cw, so $(a,b)$ and $(d,-c)$ are $90-\theta$ degrees apart.  Then $$\cos (90-\theta)= \frac{ad-bc}{\sqrt{(a^2+b^2)(c^2+d^2)}}$$  The denominator is >0, given our restriction on $\theta$, $90-\theta \in$[-90,90], and cosine is $\geq 0$ in this range.  So $ad-bc \geq 0$.
