Geometry of Young's experiment for optical path length I am currently studying the textbook Modern Optical Engineering, fourth edition, by Warren Smith. When presenting the concept of optical path length, the author says the following:

With reference to Fig. 1.13, it can be seen that, to a first approximation, the path difference between $AP$ and $BP$, which we shall represent by $\Delta$, is given by
$$\Delta = \dfrac{AB \cdot OP}{D}$$


I'm having difficulty understanding how the mathematics $\Delta = \dfrac{AB \cdot OP}{D}$ corresponds to the figure. I suspect that there is some use of trigonometry and/or geometry that I am not seeing. I would greatly appreciate it if someone would please take the time to explain this to me.
 A: Let us consider the following diagram:

When the distance $L$ between the slit plane and the screen is large compared to the distance $d$ between the slits, we can assume $S_1P$ and $S_2P$ are parallel to each other. And the $\delta$ in the image represents the optical path difference.
Now consider the light orange coloured triangle. Here $\sin \theta=\delta/d$. As $L>>d$, we can assume $\sin\theta\approx\theta$ and hence $\theta\approx\delta/d$.
Now consider the light blue coloured triangle ($POQ$). Here $\tan\theta=y/L$. Using small angle approximation we can tell $\theta\approx y/L$.
So equating $\theta\approx\delta/d$ and $\theta\approx y/L$, we get $\delta/d\approx y/L$. 
If you haven't figured it till now, I have just derived the expression in your question but used different symbols in accordance with the diagram in my answer.

Image taken from the question - In Young's double slit experiment, why are the two theta values equivalent?
A: The assumption is that AB is very small compared with D.  Then the two triangles are similar and the sine of the small angle is OP/D = Δ/AB
