Nice concurrency in rhombus

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Nice concurrency in rhombus

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#1 h Dec 15, 2012, 9:38 PM • 1 Y

Y by Adventure10

Points $X,Y,Z$ are marked on the sides $AB,BC,CD$ of the rhombus $ABCD,$ respectively, so that $XY\parallel AZ.$ Prove that $XZ,AY$ and $BD$ are concurrent.

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#2 h Dec 16, 2012, 2:37 AM • 2 Y

Y by Adventure10, Mango247

Hi!

Let me give you some advices in order to prove the property:

1- Restate your problem as follows:

Let $ABCD$ be a rhombus and take $Y\in\overline{BC}$ , $Z\in\overline{BD}$ . $\overline{AY}$ cuts $\overline{BD}$ in $P$ and $\overleftrightarrow{ZP}$ cuts $\overline{AB}$ in $X$ . Prove that $\overline{XY}\|\overline{AZ}$ .

2- $\overleftrightarrow{AD}$ touches $\overleftrightarrow{BC}$ in $H$ , so you want to prove $\dfrac{BX}{BY}=\dfrac{BA}{BH}$ right? well, this figure has two similar triangles. Can you construct new ones?

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#3 h Dec 19, 2012, 8:01 PM • 1 Y

Y by Adventure10

delegat wrote:

Points $X,Y,Z$ are marked on the sides $AB,BC,CD$ of the rhombus $ABCD,$ respectively, so that $XY\parallel AZ.$ Prove that $XZ,AY$ and $BD$ are concurrent.

$BC$ and $AZ$ intersect at $M$ and $XZ$ meets $AY$ in $O$ and $MO$ meets $AB$ in N.
Assume that $AX=a$ , $BX=b$ , $DZ=c$
Now apply Menelaus theorem so we have $XN=a^2b+ab^2-abc/ac+ab+b^2$
Assume that $BO$ meets $AZ$ in $Q$
Now apply Ceva theorem in triangle $ABM$
so we have $(BN/AN).(AQ/QM).(MY/BY)=1$ we know $MY/BY=a/b,BN/AN=2ab+a^2b+b^3/a^2c+abc$
thus $AQ/MQ=c/a+b$ also we have $BM/a+b=a+b/c$ so $AB/BM=c/a+b$
thus $BQ$ is the angle bisector of $ABM$ .
QED

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#4 h Dec 22, 2012, 7:59 PM • 2 Y

Y by Adventure10, Mango247

Well, $\Delta AZD\sim\Delta YXB$ , having parallel sides, hence Desargues...

Best regards,
sunken rock

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#5 h Dec 23, 2012, 3:54 PM • 1 Y

Y by Adventure10

Let $E \equiv XY \cap BD$ , $F \equiv AZ \cap BD$ , $P \equiv XZ \cap BD$ and $P' \equiv XZ \cap AY$ .
Then we have to prove $P \equiv P'$ .
$1.$ $\frac{PX}{PZ}=\frac{XE}{FZ}$ which is true because triangles $XEP$ and $ZFP$ are similar, due to parallel sides.

$2.$ $\frac{P'X}{P'Z}=\frac{XY}{AZ}$ similarly from triangles $XYP'$ and $ZAP'$ .

But we have $\frac{XE}{FZ}=\frac{XY}{AZ}$ and hence $\frac{P'X}{P'Z}=\frac{PX}{PZ}$ and hence since P and P' must both lie between $X$ and $Z$ we have $P \equiv P'$ $QED$

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