MHB Quadrilateral Circumscribed Circle Diagonals and Chords Concurrency Proof

[anemone]

Here is this week's POTW:

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If a quadrilateral is circumscribed about a circle, prove that its diagonals and the two chords joining the points of contact of opposite sides are all concurrent.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to Opalg for his correct solution:), which you can find below:

Spoiler

This is a special case of Brianchon's theorem in projective geometry, which says that if a hexagon is circumscribed about a conic section then its three main diagonals (those connecting opposite vertices) are concurrent.

[TIKZ]\draw circle (3cm) ;
\coordinate (A) at (-1.52,3) ;
\coordinate (B) at (5.2,3) ;
\coordinate (C) at (1.16, -4) ;
\coordinate (D) at (-4.5,-1.03) ;
\coordinate (P) at (0,3) ;
\coordinate (Q) at (2.6,-1.5) ;
\coordinate (R) at (242:3cm) ;
\coordinate (S) at (144:3cm) ;
\draw (A) node[above]{$A$} -- (B) node[above]{$B$} -- (C) node[below]{$C$} -- (D) node

{$D$} -- cycle ;
\draw (A) -- (C) ;
\draw (B) -- (D) ;
\draw (P) node[above]{$P$} -- (R) node[below]{$R$} ;
\draw [dashed] (Q) node

{$Q$} -- (S) node

{$S$} ;[/TIKZ]
In this case, where the conic is a circle, think of the quadrilateral $ABCD$ as being a hexagon with vertices $APBCRD$. Its diagonals are then $AC$, $PR$ and $BD$. Brianchon's theorem says that these three lines are concurrent. The same argument for the hexagon $ABQCDS$ shows that the lines $AC$, $BD$ and $QS$ are concurrent.​

Alternate elementary approach:

Spoiler

Any quadrilateral is a perspective projection of a parallelogram. In the preimage of this projection, which preserves intersections, the inscribed circle is an inscribed ellipse. Thus we have an ellipse inscribed in a parallelogram. The chords joining the points of contact of opposite sides go through the centre of the ellipse, and so do the diagonals of the parallelogram because the eclipse and its circumscribing parallelogram are both unchanged by rotation of half a turn about a centre, each of the two diagonals and the two chords is invariant under this transformation, and the only invariant lines are those through the centre.

Incidentally, the method shows that "circle" could be replace by "ellipse" in the problem, and so we have completed our proof.