Geometry in a square

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Geometry in a square

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#1 h May 18, 2015, 10:17 PM • 2 Y

Y by Adventure10, Mango247

Points $M$ and $N$ lie on the sides $BC$ and $CD$ of the square $ABCD,$ respectively, and $\angle MAN = 45^{\circ}$ . The circle through $A,B,C,D$ intersects $AM$ and $AN$ again at $P$ and $Q$ , respectively. Prove that $MN || PQ.$

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#2 h May 18, 2015, 10:46 PM • 3 Y

Y by sunken rock, Adventure10, Mango247

Since $\angle PAQ=\angle BAC=45^{\circ}$ $\Longrightarrow$ $PQ=BC=CD$ $\Longrightarrow$ $BPCQ$ is an isosceles trapezoid $\Longrightarrow$ $PB=QC$ and similarly $PC=QD$ $\Longrightarrow$ $\triangle PBC \cong \triangle QCD$ $\Longrightarrow$ $PBDC \cong QCAD.$ Hence if $L \equiv PD \cap BC,$ it follows that $QN:QA=PL:LD=PM:MA$ $\Longrightarrow$ $MN \parallel PQ.$

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tranquanghuy7198

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tranquanghuy7198

#4 h May 19, 2015, 4:45 AM • 1 Y

Y by Adventure10

My solution:
$S$ is the projection of $A$ on $MN$ $\Rightarrow$ $AB = AS = AD$ (familiar property)
It's easy to see that: $\triangle{APC}$ $\sim$ $\triangle{ASN}, \triangle{AQC}$ $\sim$ $\triangle{ASM}$
$\Rightarrow$ $\frac{AP}{AC} = \frac{AS}{AN}, \frac{AC}{AQ} = \frac{AM}{AS}$
$\Rightarrow$ $\frac{AP}{AQ} = \frac{AM}{AN}$ and the conclusion follows

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4381 posts

#5 h May 19, 2015, 5:40 AM • 2 Y

Y by Adventure10, Mango247

Let $O$ be the centre of the square, $K=AM\cap BD, L=AN\cap BD$ ; $\angle MAL=\angle BML=45^\circ\implies ML\bot AN$ . Similarly $NK\bot AM$ , hence $MNLK$ is cyclic, wherefrom $\angle AMN=\angle ALK\ (\ 1\ )$ .
$PC\bot AP\implies PCOK$ cyclic, so $AK\cdot AP=AO\cdot AC\ (\ 2\ )$ . Similarly $LOCQ$ is cyclic, therefore $AL\cdot AQ=AO\cdot AC\ (\ 3\ )$ . From $(2)$ and $(3)$ $PQLK$ is cyclic, so $\angle KPQ=\angle ALK\ (\ 4\ )$ ; with $(1)$ and $(4)$ we are done.
However Luis's solution is really original and effective!

Best regards,
sunken rock

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jayme

#6 h May 19, 2015, 10:43 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,

1. MLN is the Pascal’s line of APDCBQA
2. PCQD is a trapeze and PD perpendicular to AQ
3. PBQC is a trapeze and BQ perpendicular to AP
4. L is the orthocenter of APQ
5. X the symmetric of L wrt PQ
6. ML is the Pascal’s line of XCBQPAX and ML // PQ

and we are done without any calculation…

Sincerely
Jean-Louis

This post has been edited 1 time. Last edited by jayme, May 19, 2015, 10:44 AM
Reason: typo

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ENC

#7 h Jun 13, 2017, 2:54 AM • 2 Y

Y by Adventure10, Mango247

@sunken rock,
how is ML perpendicular to AN?
(how is angle AML equal to 45 degrees?)

This post has been edited 1 time. Last edited by ENC, Jun 13, 2017, 2:58 AM
Reason: typo

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PROF65

#8 h Jun 13, 2017, 11:18 AM • 1 Y

Y by Adventure10

A different approach:
Let $AM$ cuts $BD$ at $M'$ and $AN$ cuts $BD$ at $N'$ it 's obvious that $DNM'A$ and $BMN'A$ are cyclic so $\angle M'NN'=\angle M'NA = \angle M'DA=\angle BDA , \angle N'MM'=\angle N'MA \angle N'BA=\angle DBA$ but $ABD$ is isoceles thus $MNM'N'$ besides we know $M'N'PQ$ is cyclic hence Reim's yields $MN \parallel PQ$
RH HAS
differrent angle chase
Click to reveal hidden text

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Taco12

#9 h Dec 19, 2021, 1:58 AM

Y by

Let $O$ be the center of the circle. Clearly, $POQ=90^{\circ}$ . Let $\angle DAN = 90^{\circ}$ . Some angle chasing yields $\angle ANM=(x+45)^{\circ}$ , so the problem reduces to showing $\angle AQP=(x+45)^{\circ}$ . This is equivalent to showing $\angle AQO=x^{\circ}$ , since $\angle OQP = 45^{\circ}$ . Let $CD$ intersect $OQ$ at $X$ . By $AA$ , $NXQ$ is similar to $NDA$ , which means $\angle AQO = x^{\circ}$ , and the result follows.

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