76. A parallelogram and a circle (nice !).

by Virgil Nicula, Aug 3, 2010, 4:35 AM

Quote:
Let $ABCD$ be a parallelogram. Denote $O\in AC\cap BD$ and the projections $M$ , $N$ ,
$P$ of $D$ to $AB$ , $BC$ , $CA$ respectively. Prove that $OMNP$ is cyclically.
Proof. Suppose w.l.o.g. that $A<90^{\circ}$ . Observe that $\left\|\begin{array}{c}
\widehat {ODM}\equiv\widehat {OMD}\ (1)\\\
\widehat {ADM}\equiv \widehat {CDN}\ (2)\end{array}\right\|$ . Thus
$\left\|\begin{array}{ccc}
AD\parallel BC & \implies & \widehat {ADB}\equiv\widehat{DBN}\\\
BMDN-\mathrm{ cyclic} & \implies & \widehat {DBN}\equiv \widehat {DMN}\end{array}\right\|$ $\implies$ $\widehat {ADB}\equiv\widehat {DMN}$ $\implies$ $\widehat{ADM}+\widehat {ODM}=$ $\widehat{OMN}+\widehat {OMD}$ $\stackrel{(1)}{\implies}$ $\widehat{ADM}\equiv\widehat {OMN}$ $\stackrel{(2)}{\implies}$ $\widehat {CDN}\equiv\widehat {OMN}\ (3)$ .Observe that $CDPN- \mathrm{cyclic}$ $\implies$ $\widehat {CDN}\equiv\widehat{CPN}$ $\stackrel{(3)}{\implies}$ $\widehat {OMN}\equiv\widehat{CPN}$ $\implies$ $OMNP-\mathrm{cyclic}$ .

Remark. The problem from
here is a application of this problem to the parallelogram $BHCD$ .
Quote:
Proposed problem. Let $ABC$ be an acute triangle with orthocenter $H$ . Denote midpoint $M$ of $[BC]$ , $D\in AH\cap BC$ , diameter $[AA']$ in circumcircle of $\triangle ABC$ and projections $X$ , $Y$ of $H$ on $BA'$ , $CA'$ respectively. Prove that $DXYM$ is cyclically.
Proof. Apply upper problem to the parallelogram $BHCA'$ where construct perpendicular lines from $H$ on the sides of $\triangle BA'C$. See and here
Quote:
Generalization. Let $ABCD$ be a quadrilateral with $A=C\ne 90^{\circ}$ . Denote $O\in AC\cap BD$ and the projections $M$ , $N$ , $P$ of $D$ to $BC$ , $CA$ , $AB$ respectively. Prove that $OMNP$ is cyclically (problem 4435 from "Mathematika v skole", no.4/1999. p.85).

Proof (very nice Luisgeometria's proof !) If $A=C$ , then $ m(\widehat{CDM})=m(\widehat{ADP})=\alpha$ . The quadrilaterals $ADNP$ and $CDNM$ are inscribed in circles with diameters $ AD$ , $CD$ $\Longrightarrow$ $m(\widehat{CNM})=m(\widehat{CDM})=\alpha$ and $m(\widehat{ANP})=m(\widehat{ADP})=\alpha$ $\Longrightarrow$ $AC$ is an angle bisector of $\widehat{MNP}\ (*)$ . Denote the midpoints $U$ , $V$ of $AD$ , $CD$ . Thus $OU=\frac {_1}{^2}\cdot CD=VM$ and $OV=\frac {_1}{^2}\cdot AD=UP$. Since $m(\widehat{MVO})=$ $m(\widehat{PUO})=$ $m(\widehat{ADC})-2\alpha$ , it follows that $\triangle MVO \equiv\triangle OUP$ (s.a.s. criterion) , $OP=OM$ $ \Longrightarrow$ $O$ lies on the perpendicular bisector of $MP$ , then together with $(*)$ we deduce that $O$ belongs to the circumcircle of $\triangle MNP$ .
This post has been edited 15 times. Last edited by Virgil Nicula, Nov 23, 2015, 3:31 PM

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