I want to incorporate 2 diagonal lines in a logo design. The lines have to be parallel to each other and have to be exactly 0.5 inches apart when measured perpendicular. The upper point of Line 1 has to be fixed at one intersection on Point A. The lower point of Line 2 has to be fixed at a different intersection on Point D. How would I calculate Point B and Point C?
Here is one possible geometric construction, which could be done in CAD software.
Draw a circle of radius $0.5$ around $A$. Then the line $BD$ will be tangent to this circle. To find the point $E$ of tangency, note that $AED$ must be a right angle; so draw a circle with diameter $AD$, and let $E$ be where it intersects the other circle. (There are two points of intersection; choose the one on the right side.) Then $B$ will be the intersection of line $DE$ and the horizontal line through $A$. Then point $C$ can be found by completing the parallelogram.
See the modified picture below, where I have added some lines and labels.
Since triangle CKD is similar to triangle DHB, we get the proportion $$\frac{0.5}{CD}=\frac{3}{BD}$$ which gives $BD=6CD$.
Using Pythagorean theorem on triangle DHB, we get $$BH^2+9=BD^2$$ so we get $$BH^2=BD^2-9=36CD^2-9$$
Noting that $CD+CE=2.25$ and $CE=BH$, we get $CD=2.25-BH$. Continuing, we get $$BH^2=36CD^2-9=36(2.25-BH)^2-9=36(5.0625-4.5BH+BH^2)-9$$ which gives $$182.25-162BH+36BH^2-9=BH^2$$
Using $x=BH$, we need to solve $$35x^2-162x+173.25=0$$
The quadratic formula yields $$BH=CE=x=\frac{162\pm\sqrt{162^2-4(35)(173.25)}}{70}$$
This gives the two solutions 1.6772 and 2.9514. The bigger one here seems to not be possible based on the drawing. So I'd vote for 1.6772 for BH. This seems to give $AB=2.25-BH=0.5728$.
Draw a line perpendicular to line BD, emanating from point C. From trig in the resulting triangle, it follows that
$$AB=d \sin (C\hat{D}B)$$
where d is the distance between lines $AC, BD$ and from the right triangle formed by the vertical emanating from point D and the extension of segment AB
$$\sin (C\hat{D}B)=\frac{h}{\sqrt{(b+a-AB)^2+h^2}}$$
Here $h=3'', b=2'', a=0.25''$ as read from the diagram, and they represent the vertical distance, the box length and the margin respectively. Putting everything together and $AB=x$ we have to solve the quadratic equation
$$(b+a-x)^2+h^2=x^2\frac{h^2}{D^2}$$ Plugging in yields a single positive solution, $AB\approx 0.573''$.
