# 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.

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?

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)

$$AX:XZ = AY:AZ$$

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 $$2$$)

• Oh, it seems I was too late.
– ACB
Sep 17 at 14:14
• that's alright... we are all typing in answers without knowing if a similar answer was on the way and for many of these questions, it all happens within minutes :) Sep 17 at 14:21

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, $$\frac{XZ}{AX}=\frac{AZ}{AY}\implies AX:XZ=AY:AZ$$

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.

• I am a beginner so I cannot understand the meaning of few symbols used by you. Please explain the meaning of $\color{red}{(}\color{blue}{A,Z}\color{red}{;}\color{blue}{X,Y}\color{red}{)}$. Sep 17 at 16:17