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

Question

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.

1. Extend $AB$

2. 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 $c$.

3. 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.

4. 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?

Answer 1

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)

$AX:XZ = AY:AZ$

Answer 2

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 $2$)

Answer 3

Well, without additional constructions, I'll try to answer. $$\triangle ABX\sim\triangle ZDX$$ $$\begin{align}\implies\frac{XZ}{AX} & =\frac{DZ}{AB} \\ &=\frac{DC+CZ}{AB} \\ &=\frac{DC}{AB}+\frac{CZ}{AB}\\ &=1+\frac{CZ}{AB}\tag{1}\end{align}$$ Since $\triangle CYZ\sim\triangle AYB$ $$\frac{CZ}{AB}=\frac{YZ}{AY}$$ Back to (1), $$\begin{align}1+\frac{CZ}{AB} &=1+\frac{YZ}{AY} \\ &=\frac{AY+YZ}{AY}\\ &=\frac{AZ}{AY}\end{align}$$ Therefore, $$\frac{XZ}{AX}=\frac{AZ}{AY}\implies AX:XZ=AY:AZ$$

Answer 4

Another solution (killing a fly with a bazuka), using cross ratio:

\begin{align}{AX\over XZ}\cdot {YZ\over AY}&= {AX\over XZ}: {AY\over YZ}\\ &= (A,Z;X,Y)\\ & = (BA,BZ;BX,BY)\\ &= (B\infty, BZ;BD,BC)\\ &= (\infty, Z;D,C)\\ &={CZ\over DZ } \end{align}

Since $\triangle CZY\sim \triangle DZA$ we have ${CZ\over DZ }={YZ\over AZ}$ and we are done.

Answer 5

Using the Intersection Lemma we get $$\frac{AX}{XZ} = \frac{[ABD]}{[BDZ]} = \frac{AB}{DZ}$$ and $$\frac{AY}{YZ} = \frac{[ABC]}{[BCZ]} = \frac {AB}{CZ}$$ So, $\frac{AY}{AZ} = \frac{AB}{CZ+AB} = \frac{AB}{CZ +CD}$