MHB Show Equivalence of AB Parallel to XY | Wondering

[mathmari]

Hey!

We have the triangle ABC. We have that $X\in AC$, $Y\in BC$ and $Z=AY\cap BX$, where $X,Y\neq A,B,C$. I want to show that AB is parallel to XY iff $\frac{|CX|}{|CA|}=\frac{|ZB|}{|ZX|}=1$.

Could you give me a hint what we have to show?

When AB is parallel to XY then we have that then AZB and ZXY are congruent. What do we get from that? (Wondering)

 

[I like Serena]

Hey!

We have the triangle ABC. We have that $X\in AC$, $Y\in BC$ and $Z=AY\cap BX$, where $X,Y\neq A,B,C$. I want to show that AB is parallel to XY iff $\frac{|CX|}{|CA|}=\frac{|ZB|}{|ZX|}=1$.

Could you give me a hint what we have to show?

Hey mathmari! (Smile)

If we pick BX and AY to be medians, we have $\frac{|CX|}{|CA|}=\frac 12 \ne \frac 21 =\frac{|ZB|}{|ZX|} \ne 1$.
Can it be that $\frac{|CX|}{|CA|} \cdot \frac{|ZB|}{|XZ|}=1$ was intended?

When AB is parallel to XY then we have that then AZB and ZXY are congruent. What do we get from that?

Aren't $AZB$ and $ZXY$ similar instead of congruent?

Anyway, yes, I believe we should use that if $AB\parallel XY$, that then $AZB$ and $ZXY$ are similar.
That means that the corresponding sides have the same ratio.
Btw, $ABC$ and $XYC$ are also similar. (Thinking)

 

[mathmari]

If we pick BX and AY to be medians, we have $\frac{|CX|}{|CA|}=\frac 12 \ne \frac 21 =\frac{|ZB|}{|ZX|} \ne 1$.
Can it be that $\frac{|CX|}{|CA|} \cdot \frac{|ZB|}{|XZ|}=1$ was intended?

I don't know. Maybe. The exercise statement is as in #1.

Aren't $AZB$ and $ZXY$ similar instead of congruent?

Anyway, yes, I believe we should use that if $AB\parallel XY$, that then $AZB$ and $ZXY$ are similar.
That means that the corresponding sides have the same ratio.
Btw, $ABC$ and $XYC$ are also similar. (Thinking)

From the fact that the triangles $AZB$ and $ZXY$ are similar, we get that $\frac{BZ}{ZX}=\frac{AZ}{ZY}=\frac{AB}{XY}=m_1$.

From the fact that the triangles $ABC$ and $XYC$ are congruent, we get that $\frac{XY}{AB}=\frac{YC}{BC}=\frac{XC}{AC}=m_2$.

Since $m_1=\frac{AB}{XY}$ and $m_a=\frac{XY}{AB} \Rightarrow \frac{AB}{XY}=\frac{1}{m_2}$ we get that $\frac{BZ}{ZX}=\frac{AC}{XC}$, right? (Wondering)

Maybe we have to show the relation without equal to $1$ ? (Wondering)
What about the other direction? (Wondering)

Does the other direction would hold if we would have $\frac{BZ}{ZX}=\frac{AC}{XC}$ without equal to $1$ ? (Wondering)

 

[mathmari]

Does the other direction would hold if we would have $\frac{BZ}{ZX}=\frac{AC}{XC}$ without equal to $1$ ? (Wondering)

Suppose it holds that $\frac{BZ}{ZX}=\frac{AC}{XC}=:m$.

We consider the triangles $AZB$ and $XZY$. We have that $BZ=m\cdot ZX$ and the angle of $Z$ in both triangles are the same.
We consider the triangles $ABC$ and $XCY$. We have that $AC=m\cdot XC$ and the angle of $C$ in both triangles are the same.

Can we get something from here? (Wondering)

 

[I like Serena]

I don't know. Maybe. The exercise statement is as in #1.

From the fact that the triangles $AZB$ and $ZXY$ are similar, we get that $\frac{BZ}{ZX}=\frac{AZ}{ZY}=\frac{AB}{XY}=m_1$.

From the fact that the triangles $ABC$ and $XYC$ are congruent, we get that $\frac{XY}{AB}=\frac{YC}{BC}=\frac{XC}{AC}=m_2$.

Since $m_1=\frac{AB}{XY}$ and $m_a=\frac{XY}{AB} \Rightarrow \frac{AB}{XY}=\frac{1}{m_2}$ we get that $\frac{BZ}{ZX}=\frac{AC}{XC}$, right?

Yes. (Nod)

Maybe we have to show the relation without equal to $1$ ?

If so, then we also need to invert one of the fractions.
Either way, the result would be the same as replacing the $=$ sign by a multiplication.

What about the other direction?

Does the other direction would hold if we would have $\frac{BZ}{ZX}=\frac{AC}{XC}$ without equal to $1$ ?

Suppose it holds that $\frac{BZ}{ZX}=\frac{AC}{XC}=:m$.

We consider the triangles $AZB$ and $XZY$. We have that $BZ=m\cdot ZX$ and the angle of $Z$ in both triangles are the same.
We consider the triangles $ABC$ and $XCY$. We have that $AC=m\cdot XC$ and the angle of $C$ in both triangles are the same.

Can we get something from here?

I don't see how.

As I see it, either we have to assume that $XY \nparallel AB$ and deduce that the fractions are different.
Or we assume the fractions are the same and deduce that $XY \parallel AB$.
I haven't figured out how to do that yet though.
Perhaps we could do it with vectors instead of geometry properties.

 

[mathmari]

For the other direction we have that $\frac{BZ}{ZX}=\frac{AC}{XC}$.

Let $Y'\in BC$ so that $AB\parallel XY'$.
Then $Z'$ is the intersection point of $AY'$ and $BX$.
We have as in the other direction that the triangles $AZB$ and $Z'XY'$ are similar and so it holds that $\frac{BZ'}{Z'X}=\frac{AZ'}{Z'Y'}=\frac{AB}{XY'}$.
The triangles $ABC$ and $XY'C$ are also similar and so it holds that $\frac{XY'}{AB}=\frac{Y'C}{BC}=\frac{XC}{AC}$.
So, we get that $\frac{BZ'}{Z'X}=\frac{AC}{XC}$.
Since $\frac{BZ}{ZX}=\frac{AC}{XC}$, we get that $\frac{BZ'}{Z'X}=\frac{BZ}{ZX}$. Do we conclude from here that $Z'=Z$ and so that $Y'=Y$ ? (Wondering)