双心四边形OI⊥EF

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Line IO in Bicentric Quadrilaterals
Petrisor Neagoe
Maths teacher at Mathias Hammer High School, Anina, Romania, Europe
15 December, 2006
A bicentric quadrilateral ABCD is inscribed in the circle C(O,R) and circumscribes the circle C(I,r). Let E be the intersection point of the lines AB and CD and let F be the intersection point of the lines BC and AD. Prove that OI is perpendicular on EF.

users.math.uoc.gr/~pamfilos/eGallery/problems/Bicentric.html

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见forum.php?mod=viewthread&tid=6941的16#的结论2

hbghlyj

Demonstration:

Let V be the intersection point of the diagonals AC and BD. We show that: i) OC⊥EF, ii) IV⊥EF, iii) V, I, O are collinear points; and that i), ii) and iii) imply OI⊥EF.

1. Let's prove that OV⊥EF. We prove that V is the orthocenter of the ΔOEF.
   Let's prove that FV⊥EO. Let S be the intersection point of the lines FV and EO, M - midpoint of the side AB and N - midpoint of the side CD.
   We construct $V P \perp D A ( P \in D A )$, and $V Q \perp B C ( Q \in B C )$
   $\left. \begin{array} { l } { \text { M-midpoint of the side } A B \Rightarrow O M \perp A B } \\ { \text { N-midpoint of the side } C D \Rightarrow O N \perp C D } \end{array} \right\} \Rightarrow$
   $\left. \begin{array} { c } { \Rightarrow M O N E \text { - cyclic quadrilateral } \Rightarrow \measuredangle S E B \equiv \measuredangle M N O } \\\left. \array{ V P \perp A F \\ V Q \perp B F } \right\} \Rightarrow P V Q F \text { - cyclic quadrilateral } \Rightarrow \measuredangle S F B \equiv \measuredangle V P Q \end{array}\right\} \Rightarrow$
   $\Rightarrow m ( \measuredangle W F B ) - m ( \measuredangle S F B ) = m ( \measuredangle M N O ) - m ( \measuredangle V P Q ) = x$
   We construct $A A ^ { \prime } \perp B C ( A ^ { \prime } \in BC )$ and $B B ^ { \prime } \perp D A ( B ^ { \prime } \in D A )$
   Let's prove that $M P = M Q$
   $\left. \begin{array} { l } { \text { In the } \triangle M A P : M P ^ { 2 } = \frac { A B ^ { 2 } } { 4 } + A P ^ { 2 } - A B \cdot A P \cdot \operatorname { cos } A } \\ { \text { In the } \triangle M B Q: M Q ^ { 2 } = \frac { A B ^ { 2 } } { 4 } + B Q ^ { 2 } - A B \cdot B Q \cdot \operatorname { cos } B } \end{array} \right\} \Rightarrow$
   $M P = M Q \Leftrightarrow$
   $\Leftrightarrow \frac { A B ^ { 2 } } { 4 } + A P ^ { 2 } - A B \cdot A P \cdot \operatorname { cos } A = \frac { A B ^ { 2 } } { 4 } + B Q ^ { 2 } - A B \cdot B Q \cdot \operatorname { cos } B$
   $\Leftrightarrow \frac { A P } { B Q } = \frac { B Q - A B \cdot \operatorname { cos } B } { A P - A B \cdot \operatorname { cos } A }\quad$(1)
   $\triangle A P V \sim \triangle B Q V \Rightarrow \frac { A P } { B Q } = \frac { A V } { B V } \Rightarrow$
   (1)$\Leftrightarrow \frac { A V } { B V } = \frac { B Q - B A ^ { \prime } } { A P - A B ^ { \prime } } \Leftrightarrow \frac { A V } { B V } = \frac { A ^ { \prime } Q } { B ^ { \prime } P } \Leftrightarrow \frac { A V } { A ^ { \prime } Q } = \frac { B V } { B ^ { \prime } P } \Leftrightarrow \frac { C V } { C Q } = \frac { D V } { D P }$
   The last relation is true.
   Thus $MP=MQ$ and similarly $NP=NQ$
   $\Rightarrow \triangle M P N \equiv \triangle M Q N \Rightarrow M N \perp P Q$
   $x = m ( \measuredangle M N O ) - m ( \measuredangle V P Q ) = m ( \measuredangle T N D ) - 90 - 90 + m ( \measuredangle T P D ) =$
   $= m ( \measuredangle T N D ) + m ( \measuredangle T P D ) - 180 = 360 - 90 - m ( \measuredangle D ) - 180 =$
   $= 90 - m ( \measuredangle D ) \Rightarrow x = 90 - m ( \measuredangle D )$
   $m ( \measuredangle F S E ) = m ( \measuredangle B S F ) - m ( \measuredangle B S E ) = 180 - m ( \measuredangle S B F ) - m ( \measuredangle SFB) -$
   $- 180 + m ( \measuredangle S B E ) + m ( \measuredangle S E B ) = 180 - m ( \measuredangle B ) + m ( \measuredangle S E B ) - m ( \measuredangle S F B ) =$
   $= m ( \measuredangle D ) + x = m ( \measuredangle D ) + 90 - m ( \measuredangle D ) = 90$
   $\left. \begin{array} { c } { \Rightarrow m (\measuredangle F S E ) = 90 \Rightarrow F V \perp O E } \\ { \text { Analogous } \Rightarrow E V \perp O F } \end{array} \right\} \Rightarrow V \text { is the orthocenter of the } \triangle EOF \Rightarrow$
   $\Rightarrow \boxed{O V \perp E F}$
2. Let's prove that IV⊥EF.
   
   Let M, N, P, Q be the tangent points of the sides of ABCD to the circle C(I,r).
   
   According to Newton's theorem it follows that MN, PQ, AC and EF are concurent in point S, and NP, QM, BD and EF are concurent in point T. It follows that S, T, E, F are collinear. (1)
   
   
   
   
   According to Newton's theorem it follows that MP, NQ, AC and BD are concurrent in V
   $\Rightarrow M P \cap M Q = \{ V \}$
   Since MNPQ is cyclic quadrilateral, $M O \cap M Q = \{ V \}$
   $\{ S \} = I M \cap P Q$, $\{ T \} = M P \cap Q M$, and what shown in i) $\Rightarrow I V \perp S T$ (1) and (2) $\Rightarrow I V \perp E F$
3. Let's prove that V, I and O are collinear points.
   
   Let's consider $\{ G \} = A I \cap C ( O , R )$,$\{ K \} = B I \cap C ( O , R )$,$\{ H \} = C I \cap C ( O , R )$,$\{ L \} = D I \cap ( O , R )$, M-midpoint of the AC and N-midpoint of the BD.
   $\left. \begin{array} { l } { B K \text { - bisector of angle } ∡ A B C \Rightarrow K - \text { midpoint of the arc } \overparen { A D C } } \\ { M - \text { midpoint of the } A C } \\ { D L - \text { bisector of angle } ∡ A D C \Rightarrow L - \text { midpoint of the arc } \overparen{ A B C } } \end{array} \right\} \Rightarrow$
   $\Rightarrow K , M , O , L \text { - colinear points and } KL \perp AC$
   Analogous $\Rightarrow G , N , O , H - \text { colinear points and } GH \perp BD$
   Since BDKL-cyclic quadrilateral $\Rightarrow \triangle B I D \sim \triangle L I K \Rightarrow$
   $\begin{rcases}\Rightarrow \frac { B I } { L I } = \frac { B D } { L K } \Rightarrow \frac { B I } { L I } = \frac { B N } { L O }\\ N\text{ - midpoint of the }\mathrm{BD} \text { and } O \text { - midpoint of the } \mathrm{LK})\\ ∡IBN=∡ILO\end{rcases}⇒$
   $\Rightarrow \triangle I B N \sim \triangle I L O \Rightarrow ∡I N B \equiv ∡ I O L \Rightarrow ∡INV \equiv ∡ I O M\quad$(1)
   It is easy to notice that OMVN-cyclic quadrilateral $( O M \perp M V \text { and } O N \perp M V )$
   Since M-midpoint of the AC and N-midpoint of the BD, according to Newton's theorem $( O M \perp M V \text { and } O N \perp M V )$
   $\Rightarrow ∡ M N V \equiv ∡ V O M\quad$(2)
   (The center of the circle inscribed into a quadrilateral lies on the line joining the midpoints of the latter's diagonals) $\Rightarrow M , I , N\text{ - collinear points}\quad$(3)
   (1), (2) and (3) $\Rightarrow\boxed{V , I , O \text{- collinear points}}$