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?

Alternatively, you can simply use similar triangles to prove it, without any additional construction.

As $\triangle ADX \sim \triangle YBX$,
$\frac{XY}{AX} = \frac{BY}{AD} = \frac{BY}{BC} $
Adding $1$ to both sides,
$\frac{AY}{AX} = \frac{BC + BY}{BC} \tag1$
Also as $\triangle ADZ \sim \triangle YBA$,
$\frac{AZ}{AY} = \frac{AD}{BY} = \frac{BC}{BY} \tag2$
Multiplying $(1)$ and $(2)$,
$\frac{AZ}{AX} = \frac{BC+BY}{BY} \implies \frac{XZ}{AX} + 1 = \frac{BC}{BY} + 1$
So, $\frac{XZ}{AX} = \frac{BC}{BY} = \frac{AZ}{AY}$ (using $2$)
