AD and BC meet MN at P and Q

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AD and BC meet MN at P and Q

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WakeUp

#1 h Dec 28, 2010, 4:39 PM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a convex quadrilateral such that $BC=AD$ . Let $M$ and $N$ be the midpoints of $AB$ and $CD$ , respectively. The lines $AD$ and $BC$ meet the line $MN$ at $P$ and $Q$ , respectively. Prove that $CQ=DP$ .

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#2 h Dec 28, 2010, 6:02 PM • 1 Y

Y by Adventure10

Let $R,S$ be the midpoints of the diagonals $AC,BD.$ Since $BC=AD,$ then $NR=NS$ $\Longrightarrow$ Parallelogram $NRMS$ is a rhombus $\Longrightarrow$ $NM$ bisects $\angle RNS$ internally. Let $U \equiv AD \cap BC.$ Since $\angle CUD$ and $\angle RNS$ have corresponding parallel sides, it follows that their angle bisectors are parallel as well. By Menelaus' theorem for $\triangle UCD$ cut by $\overline{QPN},$ keeping in mind that $UP=UQ$ and $CN=ND,$ we get

$\frac{QU}{QC} \cdot \frac{CN}{ND} \cdot \frac{DP}{PU}=1 \Longrightarrow \ QC=DP.$

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#3 h Dec 29, 2010, 6:19 AM • 2 Y

Y by Adventure10, Mango247

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=377297 then use sine law.

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#4 h Jan 8, 2011, 2:59 PM • 2 Y

Y by Adventure10, Mango247

It is easy to see that $MN$ is parallel to the bisector of the angle $\widehat {(BC, AD)}$ , hence the triangles $\triangle MBQ$ and $\triangle AMP$ are pseudo-similar and our problem is solved then.

Best regards,
sunken rock

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#5 h Nov 4, 2020, 4:33 PM

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By Law of Sines on $\triangle MPA$ and $\triangle QMB$ , $\frac{AM}{\sin{\angle MPA}}= \frac{AP}{\sin{\angle AMP}}$ and
$\frac{BM}{\sin{\angle MQB}}= \frac{BQ}{\sin{\angle BMQ}}$ Since, $\sin{\angle AMP}=\sin{\angle BMQ}$ , we get that
$\frac{\sin{\angle MQB}}{\sin{\angle MPA}}= \frac{AP}{BQ}.$
By Law of Sines on $\triangle NPD$ and $\triangle QNC$ , $\frac{DN}{\sin{\angle NPD}}= \frac{DP}{\sin{\angle DNP}}$ and
$\frac{CN}{\sin{\angle CQN}}= \frac{CQ}{\sin{\angle CNQ}}$ Since, $\sin{\angle DNP}=\sin{\angle CNQ}$ , we get that
$\frac{\sin{\angle CQN}}{\sin{\angle NPD}}= \frac{DP}{CQ}.$ Because, $\angle MQB=\angle CQN$ and $\angle MPA=\angle NPD$ , we get that
$\frac{AP}{BQ}= \frac{DP}{CQ}\Longleftrightarrow \frac{AD+DP}{DP}= \frac{BC+CQ}{CQ}\Longleftrightarrow \frac{AD}{DP}= \frac{BC}{CQ}.$ Since $BC=AD$ , we get that $CQ=DP$ . $\quad \square$

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#6 h Nov 4, 2020, 6:32 PM • 1 Y

Y by Tudor1505

Let $S$ be the center of spiral similarity that sends $A \rightarrow B, D \rightarrow C$ , $X=AD \cap BC$ . Therefore, it's also the center of spiral similarity that sends $M \rightarrow N$ and $BS=AS$ . Hence, $\square ABSX, \square BMQS$ are cyclic quadrilaterals, also $\square PXQS$ . Hence, there is a spiral similarity center at $S$ that sends $BQ \rightarrow AP$ . But $BS=AS \implies BQ = AP \implies CQ = DP$ .

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#7 h Apr 10, 2025, 11:27 AM

Y by

Nice solution above, but it's trivial if we use Menelaus theorem on the $\triangle EBC$ , $MN$ and $\triangle EAD$ , $MN$ where $E=AB \cap CD$ .
We have: $\frac{DN}{NE} *\frac{EM}{MA} *\frac {PA}{PD}=1$ and $\frac{NC}{NE} *\frac{MB}{ME} *\frac{QB}{QC}=1$ $\implies$ $\frac{PA}{PD}=\frac{QB}{QC}$ $\implies$ $PA=QB$ .

This post has been edited 1 time. Last edited by Nari_Tom, Apr 10, 2025, 11:28 AM

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