increase in diagonal I am looking at the following figure, showing the deformation of a triangular element from time $t$ to time $t+dt$. Note that the $t$ on the left leg of the isosceles triangle simply denotes our triangle at time $t$ and the $t+dt$ on the deformed triangle means at a later time $t+dt$. 

According to the notes I am reading (page 16, eq. 28) the increase in the diagonal AC is given by
$$
\delta(AC) = \frac{a+d}{\sqrt{2}} + \frac{b+c}{\sqrt{2}}
$$
It is not clear to me why that is the case. I would just have said it should be
$$
\sqrt{(a+d)^2 + (b+c)^2}
$$
but this seems not to be the case. Does anyone see how one arrives at the first expression?

Your value for $\delta AC$ assumes that the new corner $C$ lies along the same line as the old corner $C$.  This is only the case if $(a+d)/(b+c)$ matches the slope of the old line $AC$.
As for where they got the other equation, I'm not sure.  I suspect that there is some simplification involving Taylor series at work.
