ABCD is a parallelogram. a straight line through A meets BD at X, BC at Y and DC at Z. Prove that AX:XZ = AY:AZ 
ABCD is a parallelogram. a straight line through A meets BD at X, BC at Y and DC at Z. Prove that $$AX:XZ = AY:AZ$$

My Approach
I realised that since the question seems to "data insufficient" , it has got to do something with constructions. Seeing the "ratio" I thought that it must be related with Similar Triangles.

*

*Extend $AB$


*Drop perpendiculars from points $X$,$Y$,$Z$ on $AB$. Name the points of intersection as $P$,$Q$,$R$ respectively. Call $XP$ as $a$, $YQ$ as $b$ and $ZR$ as $c$.


*I simplified the L.H.S. and R.H.S. of the required proof and obtained this expression: $$\color{blue}{\frac{1}{a} = \frac{1}{b} + \frac{1}{c}}$$ which I by no means was able to proof.


*Then I assumed $\frac{AP}{AX} = \frac{AQ}{AY} = \frac{AR}{AZ} = k$ from the property of similar triangles. $$\frac{AX}{XZ} = \frac{a}{c-a} = \frac{(1-k^2){AX}^2}{(1-k^2)({AZ}^2 - {AX}^2)}$$ $$\frac{AX}{XZ} = \frac{AX^2}{(AZ+AX)XZ}$$ which is a contradiction as $AZ ≠ 0$.
Where is my fault and how can I solve this problem?
Addendum
When I saw that the antecedent and consequent were part of the same line segment, I did not realise that it can be solved without additional construction (because if $∆ABC  \sim ∆A'B'C'$ we can write $\frac{AB}{A'B'} = \frac{BC}{B'C'}$ and since points $A$,$B$,$C$ cannot be collinear , so the terms of the ratio cannot be the part of the same straight line). Just for the sake of curiosity, I want to ask what algorithm is to be followed to find the required triangles that are to be proven similar?

I am afraid you shook me on step 3. Here is what I did with it.

Construct the line through $B$ parallel to $XZ$. Let it meet $AD$ at $K$ and $CD$ at $L$. This makes $ABLZ$ and $KAYB$ parallelograms.
$KB=AY$ and $BL=AZ$ ... (opposites sides of a parallelogram)
$AX:XZ = KB:BL$ ... (concurrent transversals cutting parallel lines)
$AX:XZ = AY:AZ$
