Collinearity in a Quadrilateral with Perpendicular Diagonals

by math_pi_rate, Sep 10, 2018, 10:04 AM

PROBLEM (Source=ayan.nmath)
In a convex quadrilateral $ABCD,$ the diagonals are perpendicular to each other and they intersect at $E.$ Let $P\ne A$ be a point on the side $AD$ such that $PE=EC.$ $\odot(BCD)\cap AD=Q\ne A.$ The circle passing through $A$ and tangent to line $EP$ at $P,$ intersects $\overline{AC}$ at $R.$ Prove synthetically that if $\angle BCD=90^{\circ}$ then $B,R,Q$ are collinear.

SOLUTION 1 (My solution): Let $\odot (BCD) \cap \odot (BER) = T$ and $CA \cap \odot (BCD) = X$. Note that $\angle BTD = 90^{\circ}$ and $\angle BTR = \angle BER = 90^{\circ}$ $\Rightarrow D,T,R$ are collinear.

Also, $BD$ is the perpendicular bisector of $CX$ $\Rightarrow -1=(B,D;C,X) \overset{R}{=} (Q,T;X,C)$.

And, By Pascal's Theorem on $BTTDQQ$, we get $BT \cap DQ$ lies on $CX$ $\Rightarrow B,A,T$ are collinear.

Thus, As $AQBE$ is cyclic, so $DA \cdot DQ = DE \cdot DB = DT \cdot DR \Rightarrow AQRT$ is cyclic $\Rightarrow \angle AQR = \angle ATD = 90^{\circ} \Rightarrow B,Q,R$ are collinear.

SOLUTION 2 (Synthetic_Potato): From power of point, we see that $EA\cdot ER = EP^2= EC^2$. Thus if $C'$ is the reflection of $C$ over $E$, we see that $EC^2=EA\cdot ER$ implies $(C,C';A,R)=-1$.
Let $R'$ be $BQ\cap CC'$. It suffices to now show that $(C,C';A,R')=-1$. Let $BA\cap \odot (CBQC'D) = A'$. Now from radical axis theorem on cyclic quadrilaterals $EAQB, EAA'D, BQA'D$, we see that $R'\equiv BQ\cap DA'\cap CC'$. Thus from brokard's theorem on $BQA'D$, the polar of $R'$ passes through $A$. From the defination of Polar, $(R',A;C',C)=-1$. Thus $R=R'$ and so, $B,Q,R$ are collinear.
This post has been edited 2 times. Last edited by math_pi_rate, Dec 31, 2018, 11:01 AM

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