collinear wanted, inscribed ABCD and perpendiculars related

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collinear wanted, inscribed ABCD and perpendiculars related

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Source: Ukraine TST 2014 p8

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#1 h May 1, 2020, 1:37 PM

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The quadrilateral $ABCD$ is inscribed in the circle $\omega$ with the center $O$ . Suppose that the angles $B$ and $C$ are obtuse and lines $AD$ and $BC$ are not parallel. Lines $AB$ and $CD$ intersect at point $E$ . Let $P$ and $R$ be the feet of the perpendiculars from the point $E$ on the lines $BC$ and $AD$ respectively. $Q$ is the intersection point of $EP$ and $AD, S$ is the intersection point of $ER$ and $BC$ . Let K be the midpoint of the segment $QS$ . Prove that the points $E, K$ , and $O$ are collinear.

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#2 h May 1, 2020, 8:44 PM • 4 Y

Y by amar_04, vuongnhattinpro, Infinityfun, Mango247

Let $X=AD \cap BC$ , and suppose $T$ is the foot of perpendicular from $X$ to $EK$ . Also let $E'$ be the inverse of $E$ in $\omega$ , and note that $EE'PRTX$ is cyclic. Since $K$ is the center of $\odot (PQRS)$ , so by Brokard's Theorem, $T$ is the inverse of $E$ in $\odot (PQRS)$ . This gives $KP^2=KT \cdot KE=KQ^2 \Rightarrow (P,R;T,E)=-1$ But then we get that $(P,R;T,E)=-1=(B,A;E'X \cap AB,E) \stackrel{X}{=} (P,R;E',E) \Rightarrow T \equiv E'$ Thus, $EK \perp E'X$ directly gives that $E,K,O$ are collinear. $\blacksquare$

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#3 h Feb 4, 2021, 2:47 PM • 1 Y

Y by Infinityfun

A synthetic solution:
Let $O_1$ and $O_2$ be the circumcenters of the triangles $EBC$ and $EAD$ respectively.
From perpendicular bisectors of line segments $BC$ and $AD$ , we have $O_1O \perp BC$ and $OO_2 \perp AD$ , which also means $EO_1 \parallel OO_2$ and $EO_2 \parallel OO_1$ .
As a result, $EO_1OO_2$ forms a parallelogram. Let $X$ denote the midpoint of $O_1O_2$ .
We see that $E,X,O$ are collinear.
After some angle chasing, we get that the (degenerate) quadrilaterals $EBSO_1$ and $EDQO_2$ are similar,so we have $\frac{EO_1}{O_1S}=\frac{EO_2}{O_2Q}$ i which gives us that $O_1O_2 \parallel QS$
As $X$ and $K$ are the midpoints of these parallel line segments $O_1O_2$ and $QS$ , we deduce that $E,X,K$ are collinear from the converse of Thales Theorem,. So, we have $E,X,K,O$ are collinear, which also implies $E,K,O$ , as the desired result.

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#4 h Jun 4, 2023, 5:27 PM

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Denote $BC \cap AD = F$ , $BD\cap AC = T_1$ , $SQ\cap PR = T_2$
Polar line of E written $(ABCD)$ is line $BT_1$ and polar line of E written $(PQRS)$ is $FT_2$ by Brocard.
If $BT_1T_2$ are collinear we are done since $EO$ and $EK$ is perpendicular to this line.

Denote $X_1$ and $Y_1$ foot of the perpendicular from $T_1$ to $BC$ and $AC$ . Similarly define $X_2$ and $Y_2$ .
We will show $\frac{T_1X_1}{T_1Y_1}=\frac{T_2X_2}{T_2Y_2}$ for the linearity but from similar triangles

$\frac{T_1X_1}{T_1Y_1}=\frac{BC}{AD}=\frac{EP}{ER}=\frac{SP}{RQ}=\frac{T_2X_2}{T_2Y_2}$ and we are done

This post has been edited 1 time. Last edited by Infinityfun, Jun 4, 2023, 5:28 PM
Reason: typo

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