Consider $ABCD$ is cyclic quadrilateral, the intersection of $AD$ and $CB$ is $E$, $AB$, $CD$ is $F$; $AC$ and $BD$ is $I$. Prove that $O$ is orthocenter of $\Delta EFI$.
Firstly, i will prove $OI\perp EF$.
Then i have $IK$ and $KF$ is bisector of $\angle BKD$ so $KF\perp IK$
But i need to prove $K;I;O$ are collinear , i am stuck here. Help me to continue and give me some another way, thanks.
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$\begingroup$Ok, say we have an inversion with respect to circle $C$ with center at $O$ and radius $r$.
Say we have an arbitrary point $P$ and let $P'$ be it image. Then line orthogonal to $PO$ through $P'$ is a polar line for (pole) $P$.
From definition we see that polar line is orthogonal to $OP$ and it is relatively easy to prove that if $A$ is on polar for $B$ then $B$ is on polar for $A$.
Also, there is a theorem (which is harder to prove, but again not so hard, just use of the power of the point) which says: however we take two points $X,Y$ from set $\{E,F,I\}$ then line $XY$ is polar for the third point.
Now you can easly prove your statement.
Edit:
Since $EF$ is polar line for $I$ we have $IO\bot EF$
Since $EI$ is polar line for $F$ we have $FO\bot EI$
Since $FI$ is polar line for $E$ we have $EO\bot FI$
So, $IO,EO$ and $FO$ are altitudes in triangle $EFI$ ...
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