Let $ ABCD$ be a convex quadrilateral. Let $ M,N$ be the midpoints of $ AB,AD$ respectively. The foot of perpendicular from $ M$ to $ CD$ is $ K$, the foot of perpendicular from $ N$ to $ BC$ is $ L$. Show that if $ AC,BD,MK,NL$ are concurrent, then $ KLMN$ is a cyclic quadrilateral.
Problem
Source: Indonesia TST 2009 First Stage Test 5 Problem 4
Tags: geometry, geometry proposed
tdl
30.12.2008 12:57
May be this problem is not exactly! Let choose $ ABCD$ so that $ \widehat{BCD}=90^0$ and $ OA=OC$ with $ O=AC\cap BD$, by Geometer Skatchpad I see that it isn't true!
Vo Duc Dien
26.01.2011 19:55
Vo Duc Dien
27.01.2011 03:13
Hint: You first have to prove that AB is perpendicular to CD; MN then parallels to the side of the triangle that has the intersection of MK and NL as its orthocenter. The rest is straightforward.
nawaites
14.07.2014 17:40
Anyone can give some proof???
XmL
14.07.2014 21:00
Let $E$ the midpoint of $AC$,and the four lines concur at $H$. Since $NH\perp (NE\parallel BC),MH\ perp (NE\parallel BC)$, hence $H$ is the orthocenter of $MNE$. Since $H,L,C,K$ are concyclic, we have $\angle NMH=\angle AEN=\angle ACD=\angle KLN$ so $M,N,K,L$ are concyclic. BTW this is also equivalent to the fact that the diagonals are perpendicular
nawaites
15.07.2014 12:05
Why angle of NMH equivalen with AEN???
nawaites
16.07.2014 10:06
Why angle of NMH equivalen with AEN???
Tsikaloudakis
16.07.2014 16:05
nawaites wrote: Why angle of NMH equivalen with AEN??? \[\begin{gathered} In\mathop {}\limits^{} tr.\mathop {}\limits^{} ADC\mathop {}\limits^{} : \hfill \\ \left. \begin{gathered} {\rm A}{\rm N} = {\rm N}D \hfill \\ AE = EC \hfill \\ \end{gathered} \right\}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} NE||DC\mathop {}\limits^{} \Rightarrow A\hat EN = E\hat CK \hfill \\ \end{gathered} \]