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A metrical relation in a trapezoid.
Virgil Nicula   4
N Jan 22, 2021 by tiny123tt
PP4. A trapezoid $ABCD\ ,\ AD\parallel BC$ with $I\in AC\cap BD\ ;\ P\in (AB)\ ,\ R\in (CD)$ so that $I\in PR\ ;\ S\in AR\cap BD$ . Prove that $CS\parallel AB\iff \frac {PB}{PA}+\frac {BC}{AD}=1$ .
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Virgil Nicula
Jun 15, 2015
tiny123tt
Jan 22, 2021
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A metrical relation in a trapezoid.
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Virgil Nicula
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Virgil Nicula
#1 h  Jun 15, 2015, 4:59 PM • 2 Y
Y by Adventure10, Mango247
PP4. A trapezoid $ABCD\ ,\ AD\parallel BC$ with $I\in AC\cap BD\ ;\ P\in (AB)\ ,\ R\in (CD)$ so that $I\in PR\ ;\ S\in AR\cap BD$ . Prove that $CS\parallel AB\iff \frac {PB}{PA}+\frac {BC}{AD}=1$ .
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Luis González
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Luis González
#2 h  Jun 15, 2015, 6:02 PM • 2 Y
Y by Adventure10, Mango247
Assume that $CS \parallel AB.$ $CS$ cuts $AD$ at $E$ and $BE$ cuts $AC,PR$ at $M,G,$ resp. $AECB$ is parallelogram with diagonal intersection $M$ $\Longrightarrow$ $M$ is midpoint of $\overline{AC}$ and from the complete $CRSI,$ it follows that $(M,B,G,E)=I(C,S,R,E)=-1$ $\Longrightarrow$ $\overline{GM}:\overline{GB}=-\overline{EM}:\overline{EB}=-1:2$ $\Longrightarrow$ $G$ is centroid of $\triangle ABC$ $\Longrightarrow$ $\tfrac{PB}{PA}+\tfrac{IC}{IA}=1$ (Cristea's theorem) $\Longrightarrow$ $\tfrac{PB}{PA}+\tfrac{BC}{AD}=1.$

The converse is taken for granted by the uniqueness of $S \in BD$ and $P \in AB.$
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drmzjoseph
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drmzjoseph
#3 h  Jun 15, 2015, 6:22 PM • 2 Y
Y by Adventure10, Mango247
$X \equiv AB \cap CD, CS \parallel AB \Longleftrightarrow (\infty,B,A,X)=(S,B,I,D)=R(A,B,P,X)$
$\Longleftrightarrow \frac{BX}{AB}=\frac{AP}{BP} \times \frac{BX}{AX} \Longleftrightarrow \frac{BP}{AP}=\frac{AB}{AX}=1-\frac{BC}{AD}$
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Virgil Nicula
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Virgil Nicula
#4 h  Jun 16, 2015, 4:01 PM • 2 Y
Y by Adventure10, Mango247
See PP4 (with my proof) from here.
This post has been edited 1 time. Last edited by Virgil Nicula, Jun 16, 2015, 4:03 PM
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tiny123tt
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tiny123tt
#5 h  Jan 22, 2021, 2:28 PM
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Luis González wrote:
Assume that $CS \parallel AB.$ $CS$ cuts $AD$ at $E$ and $BE$ cuts $AC,PR$ at $M,G,$ resp. $AECB$ is parallelogram with diagonal intersection $M$ $\Longrightarrow$ $M$ is midpoint of $\overline{AC}$ and from the complete $CRSI,$ it follows that $(M,B,G,E)=I(C,S,R,E)=-1$ $\Longrightarrow$ $\overline{GM}:\overline{GB}=-\overline{EM}:\overline{EB}=-1:2$ $\Longrightarrow$ $G$ is centroid of $\triangle ABC$ $\Longrightarrow$ $\tfrac{PB}{PA}+\tfrac{IC}{IA}=1$ (Cristea's theorem) $\Longrightarrow$ $\tfrac{PB}{PA}+\tfrac{BC}{AD}=1.$

The converse is taken for granted by the uniqueness of $S \in BD$ and $P \in AB.$

Cristea's theorem?
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