For cyclic quadrangle, prove that a line joining orthocenters is concurrent with sides Let $ABCD$ be an inscribed quadrilateral. Let $I=AC\cap BD$ and $H,K$ be orthocenters of $\Delta AID$ and $BIC$. Prove that $AB,DC$ and $HK$ are concurrent.

Here are all what i did:

*

*I proved that $SI$ is the polar of $L$ (with $(ABCD)$) where $S=AD\cap BC$.

*I proved that $HK\perp EF$ where $E,F$ are midpoints of $AB,CD$.

I don't know how to use the hypothesis $H,K$ is orthocenter of these triangle.
 A: 
In the diagram, the blue circles $(AB)$ and $(DC)$ have diameters $AB$ and $DC$ respectively.  We will show that their radical axis (red dotted line) contains the points $H,K,L$.
(*) Recall that the radical axis of two non-concentric circles is the set of points whose power with respect to the circles are equal.
(**) We'll also use the fact that the orthocenter of a triangle divides each altitude into two segments and that the product of the segment lengths is equal for all three altitudes.
$LB\cdot LA=LC\cdot LD$ because $A,B,C,D$ are on the same circle.  Hence the powers of $L$ with respect to $(AB)$ and $(DC)$ are equal, so by (*) $L$ is on the radical axis.
For $A,B,C,D$ let $A',B',C',D'$ be the feet of the respective perpendiculars to the opposite diagonal.  For example, $A'$ is the foot of the perpendicular from $A$ to diagonal $DB$. Because a right angle is subtended at $A'$ by $AB$, $A'$ is on circle $(AB)$, and similarly the other three points are on one or the other of the blue circles.
$AA'$ and $DD'$ are altitudes of $\triangle AID$, so intersect at $H$.  By (**) the powers of $H$ with respect to $(AB)$ and $(DC)$ are equal, so $H$ is on their radical axis.
Similarly, $K$ is on the radical axis.
Since the lines $AB,DC,HK$ all pass through $L$ they are concurrent.
The geometry of the complete cyclic quadrangle is very rich.  For anyone who is interested in learning more, see Clawson, The Complete Quadrilateral, Part III.
A: 
We carry out our proof in two parts. In the first part we use the chordal theorem to prove the three points $S$, $T$, and $U$ shown in $\mathrm{Fig.\space 1}$ are collinear. Desargues’ theorem is then used in the second part to show that the three lines $AB$, $CD$, and $HK$ shown in $\mathrm{Fig.\space 2}$ are concurrent.
$\underline{\text{Chordal Theorem}}:\enspace$The locus of the point, at which two given non-concentric circles possess the same power, is a straight line perpendicular to the line joining the centers of the circles. This line is known as the chordal line or the radical axis of the two circles.
$\underline{\text{Desargues’ Theorem}}:\enspace$If two triangles are perspective from a point, then they are perspective from a line as well.
$\underline{\text{Part 1}}:$
We extend the OP’s sketch by constructing the altitudes $AM$ and $DN$ of $\triangle ADI$ and the altitudes $BP$ and $CQ$ of $\triangle BIC$. Extended $AM$ and $BP$ meet at $T$. In similar vein, extended $DN$ and $CQ$ meet at $U$. Point $S$ is the point of intersection of the extended sides $DA$ and $CB$.
A circle can be drawn to go through $A$, $D$, $M$, and $N$, because $\measuredangle DMA =\measuredangle DNA = 90^o$. Since $\measuredangle BPC =\measuredangle BQC = 90^o$, $B$, $Q$, $P$, and $C$ lie on a circle. Furthermore, we can draw another circle to go thorough $C$, $Q$, $N$, and $D$, because $\measuredangle CQD =\measuredangle CND = 90^o$. It is possible to draw another circle to pass thorough $A$, $M$, $P$, and $B$, because $\measuredangle AMB =\measuredangle APB = 90^o$.
Since $A$, $B$, $C$, and $D$ lie on the given black circle, we shall write. $SA\times SD = SB\times SC$. Therefore, the powers of $S$ with respect to the two red circles are equal. Consequently, $S$ is located on the radical axis of the two red circles.
Since $C$, $Q$, $N$, and $D$ are on the green circle, we have. $UN\times UD = UQ\times UC$. This means that the powers of $U$ with respect to the two red circles are equal. So, $U$ is located on the radical axis of the two red circles as well.
Since $A$, $M$, $P$, and $B$ lie on the blue circle, we shall write. $TM\times TA = TP\times TB$. As in the case of $S$ and $U$, the powers of $T$ with respect to the two red circles are equal. Hence, $T$ also lies on the radical axis of the two red circles.
Therefore, we can state that $S$, $T$, and $U$ are collinear.
$\underline{\text{Part 2}}:$
Please pay attention to $\mathrm{Fig.\space 2}$, where the two triangles $ADH$ and $BKC$ are painted in color to put emphasis on them. Our aim is to pair off one side from each triangle such that the intersection of the two sides in each of the three possible pairs coincides with $S$ or $U$, or $T$. This can be achieved by pairing the sides $AD$, $DH$, and $HA$ of $\triangle ADH$ with the sides $CB$, $KC$, and $BK$ of $\triangle BCK$ respectively. With this paring of the sides, the vertices $A$, $D$, and $H$ of $\triangle ADH$ become the corresponding vertices to the vertices $B$, $C$, and $K$ of $\triangle BCK$ respectively.
In Part 1, we have already proved that $S$, $U$, and $T$, which happen to be the three intersections of pairs of corresponding sides of the two triangles, lie on a straight line. According to the converse of Desargues’ theorem, this is necessary and sufficient for us to state that the three straight lines $AB$, $DC$, and $HK$, which are joining the corresponding vertices of the two triangles, all meet in a point called the perspective center.
