Let points $H_C,H_B,H_D$ are foots of a perpendicular Case 1: $g$ intersect segment $BC$ in $X$ and $DC$ in $Y$. Then $\triangle ABX \sim \triangle YCX \sim \triangle YDA$ and so $\frac{H_D-H_B}{H_C}=\frac{AD-XB}{XC}=\frac{BC-XB}{XC}=1$ Case 2: $g$ intersect segment $CD$ - same way as case 1. Case 3.$g$ does not intersect segment $BC$ but $g$ intersect line $BC$ in $X$. Then $\triangle H_CXC \sim \triangle H_BXB \sim \triangle H_DAD$ and so $\frac{H_D+H_B}{H_C}=\frac{XB+AD}{XC}=\frac{XC}{XC}=1$ Case 4. $g$ does not intersect segment $CD$ but $g$ intersect line $CD$. Same as case 4

I think you just have to consider two cases: $g$ passes and doesn't pass the parallelogram.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20V.pdf p. 55... Sincerely Jean-Louis

This problem can be solved solely using triangle similarity. I've posted my solution HERE in case anyone is interested, as well as many other geometry materials.

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Analytic put $A(0,0),B(a,0),C(a+b,c),D(b,c)$ and let the gradient of the line passing through A is $t$ or it is constant line which is 0 thus the equation of $Y_g = tx$, $Y_g=0$ for 0 is quite easy B to g $0$ D to g is $|c|$ C to G is also $|c|$ $|0+|c|| = |c|$ $|0-|c||=|c|$ for $y = tx$ distance between point and line is $\frac{|a_{x_1}+b_{y_1}+c|}{\sqrt{a^2+b^2}}$ from B to g $ \frac{|ta|}{\sqrt{t^2+1}}$ from D to g $\frac{|tb-c|}{\sqrt{t^2+1}}$ from C to g $\frac{|t(a+b)-c|}{\sqrt{t^2+1}}$ thus we need only to proof $||ta|+|tb-c|| = |t(a+b)-c|$ after squaring which lead us to $|t^2ab-tac|=t^2ab-tac$ for $t^2ab \geq tac$ and $||ta|-|tb-c|| = |t(a+b)-c|$ after squaring $tac-t^2ab = 2|t^2ab-tac|$ for $tac\geq t^2ab$ done

Let $g$ be the $x$ axis, $A = (0,0), D =(d,e), B=(b,f)$ and $C = (b+d,e+f)$. The distance from $C$ to $g$ is $|e+f|$, distance from $B$ to $g$ is $|f|$ and distance from $D$ to $g$ is $|e|$. Because $|e+f|$ is either $|e|+|f|$ or $||f|-|e||$, we are done