MEMO 2009 pojedinačno problem 3
Dodao/la:
arhiva28. travnja 2012. Let
![ABCD](/media/m/9/c/e/9ce25711ba18d9663b73c3580de4bf5a.png)
be a convex quadrilateral such that
![AB](/media/m/5/2/9/5298bd9e7bc202ac21c423e51da3758e.png)
and
![CD](/media/m/8/9/5/895081147290365ccae028796608097d.png)
are not parallel and
![AB=CD](/media/m/a/6/0/a606bbec28ac3033e2a28d40df330502.png)
. The midpoints of the diagonals
![AC](/media/m/6/4/7/647ef3a5d68f07d59d84afe03a9dc655.png)
and
![BD](/media/m/1/1/f/11f65a804e5c922ee28a53b1df04d138.png)
are
![E](/media/m/8/b/0/8b01e755d2253cb9a52f9e451d89ec11.png)
and
![F](/media/m/3/e/8/3e8bad5df716d332365fca76f53c1743.png)
, respectively. The line
![EF](/media/m/f/5/5/f5594d5ec47ea777267cf010e788fedd.png)
meets segments
![AB](/media/m/5/2/9/5298bd9e7bc202ac21c423e51da3758e.png)
and
![CD](/media/m/8/9/5/895081147290365ccae028796608097d.png)
at
![G](/media/m/f/e/b/feb7f8fc95cee3c3a479382202e06a86.png)
and
![H](/media/m/4/c/0/4c0872a89da410a25f00b86366efece7.png)
, respectively. Show that
![\angle AGH = \angle DHG](/media/m/2/7/d/27d78dabc4afd5693c4fec5b3decb70f.png)
.
%V0
Let $ABCD$ be a convex quadrilateral such that $AB$ and $CD$ are not parallel and $AB=CD$. The midpoints of the diagonals $AC$ and $BD$ are $E$ and $F$, respectively. The line $EF$ meets segments $AB$ and $CD$ at $G$ and $H$, respectively. Show that $\angle AGH = \angle DHG$.
Izvor: Srednjoeuropska matematička olimpijada 2009, pojedinačno natjecanje, problem 3