anglegeometryplane-geometrytrianglestrigonometry
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?
The eleven postulates are sufficient to prove 3.11.
Lemma 1 A line and a point not on it, two different lines in a plane, or two parallel lines define a plane.
Two points on a line and a point not on it define a plane by #7. If two lines are different there's a point on the second that's not on the first (by #6), so by the first part they define a plane. By definition two parallel lines are different lines in a plane so define it by the second part.
Lemma 2 If $a,b,t$ are different coplanar lines and $a$ is parallel to $b$ and $t$ is not parallel to $a$ then $t$ is a transversal of $a$ and $b$.
By definition $t$ intersects $a$ so call the point of intersection $A$ defining an angle $\angle at\ne 0$ (by #3). Let $S$ be a point on $b$ then $SA$ defines a line $s$ (by #6) which is a transversal of $a$ and $b$ (by definition). Then $s$ cuts off angles $\angle sb=\angle sa$ (by #10) and $\angle st\ne \angle sa$ (by #4 because they are coincident), so $t$ is not parallel to $b$ by $\angle st\ne \angle sb$ and #10, and is a transversal (by definition).
Proposition If $a,b,c$ are different lines with $a$ parallel to $b$ and $b$ parallel to $c$ then $a$ is parallel to $c$.
If the lines are coplanar then let $t$ be a line intersecting $b$, then applying Lemma 2 twice it is a common traversal of $a,b,c$. By #10 $\angle ta=\angle tb=\angle tc$ and by #11 $a$ is parallel to $c$.
If the lines are not coplanar, then let $C$ be a point on $c$. By Lemma 1 $a$ and $b$ are in a plane $\pi_1$, $b$ and $c$ are in a different plane $\pi_2$, and $a$ and $C$ are in a plane $\pi_3$. By #9 $\pi_2$ and $\pi_3$ intersect in a line $l$ that contains $C$.
$l$ cannot intersect $b$ in any point $B$, otherwise $a$ and $B$ are in both $\pi_1$ and $\pi_3$, so $\pi_1\equiv\pi_3$ by Lemma 1, $b\equiv \pi_1\cap\pi_2\equiv\pi_3\cap\pi_2\equiv l$ which would require $C$ to be on $b$, contradicting that $b$ and $c$ are parallel. So $l$ does not intersect $b$ but it intersects $c$ at $C$. Since $b,c,l$ are coplanar in $\pi_2$, by Lemma 2 they cannot all be different, so $l\equiv c$.
$l$ cannot intersect $a$ in any point $A$, otherwise $b$ and $A$ are in both $\pi_1$ and $\pi_2$, so $\pi_1\equiv\pi_2$ by Lemma 1, contradicting that $a,b,c$ are not coplanar. Since $l\equiv c$ and $a$ are both in $\pi_3$ and do not intersect it follows that $a$ is parallel to $c$.
First, a bit of background so that others also understand what this is about.
We have a non-degenerate triangle (one that has nonzero area, so is not a point or a line) with vertices $\vec{p}_1$, $\vec{p}_2$, and $\vec{p}_3$. Any point $\vec{p}$ within this triangle can be specified via barycentric coordinates $u$, $v$, $w$: $$\begin{aligned} \vec{p} &= u \vec{p}_1 + v \vec{p}_2 + w \vec{p}_3 \\ u + v + w &= 1 \\ 0 &\le u \le 1 \\ 0 &\le v \le 1 \\ 0 &\le w \le 1 \\ \end{aligned}$$ This also implies that $w = 1 - u - v$, i.e. $$\vec{p} = u \vec{p}_1 + v \vec{p}_2 + (1 - u - v) \vec{p}_3$$ Reordering the terms above we can also write it equivalently as $$\vec{p} = \vec{p}_3 + u (\vec{p}_1 - \vec{p}_3) + v (\vec{p}_2 - \vec{p}_3)$$ which is equivalent to a linear (but not necessarily orthogonal) coordinate system in $u$,$v$ with origin at $\vec{p}_3$, $u$ axis $\vec{p}_1-\vec{p}_3$, and $v$ axis $\vec{p}_2-\vec{p}_3$. This, in turn, means that the transformation from barycentric coordinate system $\overline{b} = (u, v)$ (with $0 \le u, v, w \le 1$ and $w = 1 - u - v$) to Euclidean 2D coordinate system can be described as $$\begin{aligned} \vec{p} &= \vec{p}_3 + \mathbf{M} \overline{b} \\ \mathbf{M} &= \left[ \begin{matrix} \vec{p}_1 - \vec{p}_3 & \vec{p}_2 - \vec{p}_3 \end{matrix} \right ] \\ \end{aligned}$$ (Note, $\mathbf{M}$ is a matrix with 2 columns, and as many rows as there are dimensions in the Euclidean coordinate system.)
With this, we have accidentally proven that the transform between Euclidean coordinates and barycentric coordinates is an affine transformation, because it is the composition of two functions: a translation and a linear map.
By definition, affine transformations preserve lines and parallelism.
In two dimensions, the inverse mapping is $$\overline{b} = \mathbf{M}^{-1} \left(\vec{p} - \vec{p}_3\right)$$ where $$\mathbf{M} = \left[\begin{matrix} m_{11} & m_{12} \\ m_{21} & m_{22} \end{matrix}\right] \quad \iff \quad \mathbf{M}^{-1} = \left[\begin{matrix} \frac{m_{22}}{m_{11} m_{22} - m_{12} m_{21}} & \frac{- m_{12}}{m_{11} m_{22} - m_{12} m_{21}} \\ \frac{- m_{21}}{m_{11} m_{22} - m_{12} m_{21}} & \frac{m_{11}}{m_{11} m_{22} - m_{12} m_{21}} \end{matrix}\right]$$ In three dimensions, the easiest is to check which component in $(\vec{p}_1 - \vec{p}_3)\times(\vec{p}_2 - \vec{p}_3)$ is the largest in magnitude, and drop that Euclidean coordinate component, and do the above transform using only the two coordinate axes left. It is guaranteed to provide exact correct results, if the triangle is nondegenerate.
If the triangle is degenerate, then the common denominator is zero, $m_{11} m_{22} - m_{12} m_{21} = 0$, and no inverse exists.
There are many different ways to describe lines in barycentric coordinates. Remember, that although there are three coordinates, one of them can always be trivially eliminated ($w = 1 - u - v$), and is usually omitted, especially in computer programs and simulations using barycentric coordinates. (It can be confusing to see barycentric coordinates with only two components, but one only needs to remember that the third is always related to the two via $w = 1 - u - v$, because full barycentric coordinates sum to 1.)
(If we allow full real range in $u$ and $v$ and $w$, with $w = 1 - u - v$, the barycentric coordinates can cover the full plane the non-degenerate triangle is in. This means that if we know the triangle has nonzero area – is not a point or a line –, we can safely use $u$, $v$ coordinates anywhere on the plane, not only inside the triangle. It is then useful to remember that the triangle edges are at $u = 0$, $v = 0$, and $w = 0$ i.e. $u + v = 1$.)
However, affine transforms do not necessarily conserve distance. Even though the distance between the two lines were to stay constant in Euclidean coordinates, just rotating them can change their distance in barycentric coordinates. This is because barycentric coordinates are not orthonormal.
Therefore, I would recommend describing each line in the barycentric coordinates using two points. (This means that if you actually have a point and a direction, you can use the point+direction as the second point, even though it may be outside the triangle.) Then, transform the four points back to Euclidean coordinates. Convert back to point ($\vec{a}_0$ for one line, $\vec{b}_0$ for the other) and direction (difference between the two points, $\vec{a}_\Delta$ for one line, and $\vec{b}_\Delta$ for the other).
The cosine of the angle $\varphi$ between the two lines is $$\cos\varphi = \frac{\left\lvert \vec{a}_\Delta \cdot \vec{b}_\Delta \right\rvert}{\left\lVert\vec{a}_\Delta\right\rVert ~ \left\lVert\vec{b}_\Delta\right\rVert}$$ We take the absolute value of the dot product, because we consider line $\vec{a}_0$,$\vec{a}_\Delta$ the same as $\vec{a}_0$,$\vec{a}_\Delta$. If the cosine is close enough to 1, you can consider them parallel.
When the two lines are parallel, we can solve their distance $d$ by using a temporary variable $t$ so that $\vec{a}_0 + t \vec{a}_\Delta$ is at minimum distance to $\vec{b}_0$. This occurs when their distance reaches an extremum, and the solution is $$t = \frac{\vec{a}_\Delta \cdot \left( \vec{b}_0 - \vec{a}_0 \right) }{\left\lVert \vec{a}_\Delta \right\rVert^2}$$ Then, $$d = \left\lVert \vec{a}_0 + t \vec{a}_\Delta - \vec{b}_0 \right\rVert$$
Depending on how you describe your lines, it is usually possible to combine the above steps, so that fewer actual arithmetic operations are needed as plain reading of the above would indicate. That is, I didn't write the above to be "optimum" in any way; I tried to make it as easy to understand as possible.
I recommend you use a Computer Algebra System such as free Maxima (wxMaxima is the graphical user interfaced one) available for all operating systems to explore the details. In particular, if the direction vectors have unit lengths in Euclidean coordinates, you'll find a lot of the math related to lines becomes much, much simpler. (You'll be able to omit many $\lVert\vec{a}_\Delta\rVert$ and $\lVert\vec{b}_\Delta\rVert$ multipliers and divisors, in particular.)
Best Answer
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.