Similar triangles and parallelism

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Source: OMEC Ecuador National Olympiad Final Round 2022 N3 P5 day 2

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KAME06

#1 h Nov 4, 2024, 8:26 PM

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Let $ABC$ be a 90-degree triangle with hypotenuse $BC$ . Let $D$ and $E$ distinct points on segment $BC$ and $P, Q$ be the foot of the perpendicular from $D$ to $AB$ and $E$ to $AC$ , respectively. $DP$ and $\text{[math]}$ intersect at $R$ .
Lines $CR$ and $AB$ intersect at $M$ and lines $BR$ and $AC$ intersect at $N$ .
Prove that $MN \parallel BC$ if and only if $BD=CE$ .

This post has been edited 1 time. Last edited by KAME06, Nov 4, 2024, 8:26 PM

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Ianis

#2 h Nov 5, 2024, 1:38 AM • 1 Y

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This holds for an arbitrary triangle $ABC$ , provided that you define $R$ as the point such that $DR\parallel CA$ and $ER\parallel AB$ .

Let $O=AR\cap BC$ . By Ceva and Thales we get that $MN\parallel BC$ if and only if $O$ is the midpoint of $BC$ . Since $ABC$ and $RDE$ are homothetic with centre $O$ , we get that $O$ is the midpoint of $BC$ if and only if it is the midpoint of $BC$ and $DE$ , which happens if and only if $BD=CE$ .

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#3 h Apr 3, 2025, 7:41 AM

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Part 1: $CE=BD$ $\Rightarrow MN\parallel BC$ .
Let $O$ circumcentre of $\triangle ABC$
Let $O_1$ circumcentre of $\triangle DRE$

Claim 1: $O\equiv O_1$
As $CE=BD \iff CO+OE=DO+OB \iff OE=DO$ . It’s clearly to see that in $\triangle DRE$ , $\angle DER=90^\circ$ by problem conditions. So we have the fact $O_1E=O_1D$ , then it´s enough for $O\equiv O_1$ $\square$

Claim 2: $A, R$ and $O$ are collinear.
By Claim 1, we have $\angle ODR+\angle DRO=2\angle ODR=\angle ROE$ because $DO=OR$ , notice that $AC\parallel DP$ then $\angle ODR=\angle OCA=\angle CAO$ because $CO=OA$ . Then we have $\angle OCA+\angle CAO=2\angle ODR=\angle AOE$ . Then, $\angle ROE=\angle AOE$ means that $A, R$ and $O$ are collinear $\square$

By Claim 2, Applying Ceva´s Theorem: $\frac{AN}{NC} \cdot \frac{CB}{BO} \cdot \frac{OR}{AR}=1=\frac{AM}{MB} \cdot \frac{CB}{CO} \cdot \frac{OR}{AR} \iff \frac{AN}{NC}=\frac{AM}{MB} \iff \frac{NC}{AN}=\frac{MB}{AM} \iff \frac{AC-AN}{AN}=\frac{AB-AM}{AM} \iff \frac{AC}{AN}-1=\frac{AB}{AM}-1 \iff \frac{AC}{AN}=\frac{AB}{AM}$
by thales we concluse $MN\parallel BC$ where we complete the part 1.

Clearly For part 2: $MN\parallel BC \Rightarrow CE=BD$ are symmilar back to front of part 1. So we´re done.

This post has been edited 5 times. Last edited by jordiejoh, Apr 4, 2025, 2:02 AM

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