Collinearity in cyclic quadrilateral

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Collinearity in cyclic quadrilateral

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#1 h Dec 2, 2006, 1:05 PM • 2 Y

Y by Adventure10, Mango247

A convex quadrilateral $ABCD$ with $AC \neq BD$ is inscribed in a circle with center $O$ . Let $E$ be the intersection of diagonals $AC$ and $BD$ . If $P$ is a point inside $ABCD$ such that $\angle PAB+\angle PCB=\angle PBC+\angle PDC=90^\circ$ , prove that $O$ , $P$ and $E$ are collinear.

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#2 h Dec 2, 2006, 2:28 PM • 2 Y

Y by Adventure10, Mango247

Let $\Sigma, \Sigma_{1}, \Sigma_{2}$ be respectively the circumcircles of $ABCD, \triangle PBD, \triangle PAC$ . Let also $C_{1}, C_{2}$ be the centres of $\Sigma_{1}, \Sigma_{2}$ respectively.

Consider the radical axes of each two of them. Since the radical axis of $\Sigma, \Sigma_{1}$ is $BD$ and that of $\Sigma, \Sigma_{2}$ is $AC$ , the radical axis of $\Sigma_{1}, \Sigma_{2}$ must be concurrent with $AC$ and $BD$ , i.e. it passes through $E$ . Therefore, the radical axis of $\Sigma_{1}, \Sigma_{2}$ is $PE$ .

To prove $O$ , $P$ , $E$ are collinear, it suffices to prove $O$ has the same power with respect to $\Sigma_{1}$ and $\Sigma_{2}$ .

Without loss of generality we may let $\angle A$ and $\angle B$ to be not smaller than a right angle. Then point $P$ lies in $\triangle CDE$ and it is outside $\Sigma_{1}$ and $\Sigma_{2}$ . By some angle chasing we can prove that $\angle ODC_{1}=90^\circ$ , and so the power of $O$ w.r.t $\Sigma_{1}$ is equal to $OC_{1}^{2}-CC_{1}^{2}=OC^{2}$ . Similarly, the power of $O$ w.r.t. $\Sigma_{2}$ is also the square of the circumradius of $ABCD$ . The result follows.

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Megus

#3 h Dec 2, 2006, 7:26 PM • 2 Y

Y by Adventure10, Mango247

Consider inversion with center $P$ and any power. Then see that our hypothesis implies that $A^{*}B^{*}C^{*}D^{*}$ ( $X^{*}$ denotes point $X$ after inversion) is a rectangle. Let $O'$ be center of circumcircle of $A^{*}B^{*}C^{*}D^{*}$ . Point $E^{*}$ is the intersection of circumcircles of $PBD$ and $PCA$ and hence lies on line $PO'$ (as $O'$ is the intersection of $A^{*}C^{*}$ and $B^{*}D^{*}$ ). But well-known fact says that $P,O',O$ are collinear and hence $P,O',O^{*}$ are collinear, so $P,E^{*},O^{*}$ are collinear and hence $P,E,O$ are collinear.

QED

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#4 h Dec 3, 2006, 1:10 AM • 2 Y

Y by Adventure10, Mango247

Let $A', B', C', D'$ be the intersection point of the circle and $AP, BP, CP, DP$ respectively. By angle tracing, $A'B'C'D'$ is a rectangle. Therefore $O$ is the intersection point of $A'C'$ and $B'D'$ . Use central projection to map $P$ to the centre of the circle. Then by symmetry, $P, E, O$ are collinear.

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#5 h Dec 3, 2006, 7:33 AM • 2 Y

Y by Adventure10, Mango247

Diagram:
hope that helps a little (It helped me...)

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vslmat

#6 h Aug 5, 2012, 11:25 AM • 2 Y

Y by Adventure10, Mango247

Another solution:
Let $AP$ cut the circle at $F$ , $BP$ cut the circle at $H$ .
$\angle PAB + \angle PCB = \angle PCF = 90^{\circ}$ ,so $PC$ must cut $OF$ at a point $G$ on the circle.
Similarly, $DP$ must cut $OH$ at $K$ on the circle.
Let $GB$ cut $AK$ at $X$ .
By Pascal theorem in $ACGBDK: E = AC \cap BD, P = CG \cap DK, X = AK \cap GB$ are collinear.
By Pascal theorem in $GFAKHB: O = GF \cap KH, P = FA \cap HB, X = GB \cap AK$ are collinear.
So $E, P, X, O$ are collinear

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#7 h Oct 29, 2015, 5:53 AM • 4 Y

Y by FabrizioFelen, Pirkuliyev Rovsen, Adventure10, Mango247

If $AP,BP,CP,DP$ cut to $\odot (ABCD)$ at $A',B',C',D'$ respectively. Consider $\mathcal{G}$ the composition of involutions with poles $P,E,P$ that fixed the conic $\odot (ABCD)$ is well-known that is a involution with pole on $EP$ . Since $\mathcal{G}(A')=C'$ and $\mathcal{G}(B')=D' \Rightarrow A'C'$ and $B'D'$ are secants at the pole of $\mathcal{G}$ . Using the condition $\angle PAB+\angle PCB=\angle PBC+\angle PDC=90^\circ$ we get $A'C'$ and $B'D'$ , and are secants at $O$ so $E,P,O$ are collinear.

This post has been edited 1 time. Last edited by drmzjoseph, Oct 29, 2015, 7:57 AM

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#8 h Oct 29, 2015, 8:40 AM • 2 Y

Y by Adventure10, Mango247

This was in Korean Postal Coaching..
Let $O_1$ be the pole of $AC$ with respect to $O$ .
Draw a circle - with $O_1$ as its center and $O_1A=O_1C$ as its radius.
Then we have $2\angle APC = 2(90+\angle ADC)=180+\angle AOC = 360-\angle AO_1C$ , so $P$ lies on the circle $O_1$ .
Similarly, $P$ lies on the circle $O_2$ . I claim that $O, P, E$ lie on the radical axis of $O_1, O_2$ .
First, $O$ lies on the radical axis since $P_{O_1}(O)=OA^2=OB^2=P_{O_2}(O)$ .
Also, $E$ lies on the radical axis since $AE \cdot EC = BE \cdot ED$ . Therefore, we are done. $\blacksquare$

This post has been edited 1 time. Last edited by rkm0959, Oct 29, 2015, 8:40 AM

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