A convex quadrilateral, bisectors, circumcircles..

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A convex quadrilateral, bisectors, circumcircles..

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power of a point

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Source: Tuymaada 2001, day 2, problem 3.

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#1 h Apr 30, 2007, 8:07 PM • 2 Y

Y by Adventure10, Mango247

$ABCD$ is a convex quadrilateral; half-lines $DA$ and $CB$ meet at point $Q$ ; half-lines $BA$ and $CD$ meet at point $P$ . It is known that $\angle AQB=\angle APD$ . The bisector of angle $\angle AQB$ meets the sides $AB$ and $CD$ of the quadrilateral at points $X$ and $Y$ , respectively; the bisector of angle $\angle APD$ meets the sides $AD$ and $BC$ at points $Z$ and $T$ , respectively.
The circumcircles of triangles $ZQT$ and $XPY$ meet at point $K$ inside the quadrilateral.
Prove that $K$ lies on the diagonal $AC$ .

Proposed by S. Berlov

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#2 h Apr 30, 2007, 8:55 PM • 2 Y

Y by Adventure10 and 1 other user

First we can see that in fact $AC$ is the radical axis of the circles $PXY$ and $QZT$ , therefore their intersection in the interior of the quadrilateral will trivially lie on $AC$ .

Now, let's prove that $AC$ is their axis. We will prove only that $A \in d$ , where $d$ is the radical axis, for $C$ beeing the same thing.

But $A \in d \iff AZ \cdot AQ = AX \cdot AP$ , (the power of $A$ w.r.t to the circles), but we have that $AD \cdot AQ = AB \cdot AP$ , because the quadrilateral $BPQD$ is cyclic, therefore we only need to prove that :

$\frac {AD}{AZ} = \frac {AB}{AX}\iff \frac {PD}{AP} = \frac {AQ}{QB}$

$\iff \frac {PD}{AQ} = \frac {AP}{QB}$ , which is obvious because the triangles $\triangle{PAD}$ and $\triangle{QAB}$ are similar.

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#3 h Oct 18, 2019, 12:03 AM • 1 Y

Y by Adventure10

Sorry for bumping such an old post, but what exactly are "half-lines" ?

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#4 h Oct 18, 2019, 6:51 AM • 1 Y

Y by Adventure10

Rays @above

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