Trapezoid: what is the shortest distance between the shorter base and the line parallel to it?

Question

I want to find the length of z segment such that the red line bisects the blue line.

Note:

• blue line doesn't necessarily bisect red line and the two sides touching its end

• green line (z) is included in the measurement of the left side

Answer 1

Label the trapezium anticlockwise starting on the top left as $ABCD$, and let the blue line meet $AB$ at $E$, $CD$ at $F$, and the red line at $G$. Now, it's fairly easy to see by angle chasing that we have four similar triangles, namely $$ \triangle DCA, \quad \triangle FCG, \quad \triangle ABC, \quad \triangle AEG. $$ We know the lengths of $AB,AD,DC,BC,AC$. Also worth noting is that it so happens that $\angle ACD$ is a right-angle, but we would still expect the problem to be solvable if it weren't (move $D$ backwards and forwards: this changes $z$, but we wouldn't expect it to change our ability to find $z$), so we don't expect to need the triangles on the left to be similar to those on the right.

$\triangle ABC, \triangle AEG$ are similar, so $$ \frac{a}{AB-z} = \frac{BC}{AB}, $$ which is one equation for $a$ and $z$. Now we need to find $a$. $\triangle DCA, \quad \triangle FCG$ are similar, so $$ \frac{AD}{AC} = \frac{a}{GC} = \frac{a}{AC-AG}. $$ Another equation is given by $$ \frac{AC}{BC} = \frac{AG}{a}, $$ using again the similarity of $\triangle ABC$ and $\triangle AEG$. Solving all these simultaneously to eliminate $AG$ and then $a$, and we find that (I believe) $$ z = \frac{AB \cdot BC}{AD+BC}. $$ Is this plausible? For $BC$ very small compared to $AD$, we would expect $z$ to be small, which it is. For $BC$ much bigger than $AD$, $z$ is very close to $AB$. And the diagram is entirely determined by these three lengths, so it makes sense.

Answer 2

Consider the two triangles determined by the diagonal (red line), and then the portions of them having $a$ as a base.
Each portion is similar to the whole. So for each you can establish a proportion between bases and height:
- $a/b = (h-z)/h$
- $a/B = z/h$