This problem be constructed from 2 lemma (I think it' very useful): Lemma 1(I think it' very useful): Let $ ABCD$ is a cyclyc quadrilatel in $ (O)$. $ AB$ intersects $ CD$ at $ E$, $ AD$ intersects $ BC$ at $ F$, and $ AC$ intersects $ BD$ at $ R$. We have: a) $ P_{E}^{2}+P_{F}^{2}= EF^{2}$ b) $ R$ is orthorcenter of $ \Delta OEF$. Lemma 2 Let $ ABCD$ is a cyclyc quadrilatel in $ (O)$. $ AB$ intersects $ CD$ at $ E$, $ AD$ intersects $ BC$ at $ F$, the tangent at $ A$ and $ C$ intersects each other at $ M$, the tangent at $ B$ and $ D$ intersects each other at $ N$. We have: $ M, N, E, F$ are conlinear. (This lemma can prove by easy way if we use Pascal theory) Solution for the problem: Call the intersection of $ (O)$ with $ AB, BC, CD, DA$ are $ M, N, P, Q$ respectively. Use two lemma we get the result.

I think it is easy when I saw this problem I solved it in ten minutes I saw a very easy problem in last January(or before this exam): Let ABCD be a quadrilateral and a circle W is inscribed in it.Let K is the tangent point of AB to W,L is the tangent point of BC to W,M is the tangent point of CD to W,N is the tangent point of DA to W.Show that AC,BD,KM and LN are concurrent. In the problem of Turkey TST this problem and inversion finish the proof.

I think I saw Darij post this before using something called Newton's theorem. It says that if W,X,Y, and Z are the points of tangency of the circle to AB, BC, CD, and DA, then K is also the intersection of WY and XZ. Anyways, if this is true then ZX is the polar of Q wrt O and WY is the polar of P wrt O so since K is the intersection of both polars, PQ is the polar of K wrt O, which completes the proof. EDIT: 1000th post! Also, a nice proof is given here http://www.cut-the-knot.org/Curriculum/Geometry/CircumQuadri.shtml.

I solved with trigonometry Let $ N$ is tangent point of $ DC$ to $ K$ Let $ M$ is tangent point of $ AD$ to $ K$ Let $ L$ is tangent point of $ BC$ to $ K$ Let $ W$ is tangent point of $ AB$ to $ K$ We know $ ML - WN - BD - AC$ are concurrent; $ \angle OPC = \angle OPW = \angle ONW = \angle OWN = \alpha$ $ \angle MQO = \angle OQL = \angle OLM = \beta$ $ \angle PQC = x,\angle QPC = y$ Let $ OX\perp PQ$ $ \implies$ $ |OX| = \frac {rsin(\alpha + y)}{sin\alpha}$ $ \angle EON = \theta$ $ \implies$ $ |OE| = \frac {rsin\alpha}{sin(\alpha + \theta)}$ in $ \Delta AQP$; $ sin\beta sin(\alpha + y) = sin\alpha sin(\beta + x)$ in $ \Delta NOL$ ; $ sin\beta sin(\alpha + \theta) = sin \alpha sin(x + y - \theta + \beta)$ $ \implies$ $ y = \theta$ $ \implies$ $ |OX|.|OE| = r^{2}$ $ \blacksquare$

A step by step proof Let the circle touch $AB$, $BC$, $CD$, $DA$ at $S$, $N$, $M$, $L$, respectively. Let $R$ be the intersection of $LN$ ile $SM$. Sine Theorem for $\triangle RDM$ and $\triangle RLD$ gives \[ \begin{align} \dfrac{RD}{DM} = \dfrac{\sin \angle RMD}{\sin \angle DRM} = \dfrac{\sin \angle RLD}{\sin \angle LRD} \tag {1} \end {align*}\] Sine Theorem for $\triangle RBN$ and $\triangle RBS$ gives \[\begin{align} \dfrac{RB}{BN} = \dfrac{\sin \angle RNB}{\sin \angle BRN} = \dfrac{\sin \angle RSB}{\sin \angle SRB} \tag {2} \end {align*}\] Since $\sin \angle RMD = \sin \angle RSB$ and $\sin \angle RLD = \sin \angle RNB$, after multiplying $(1)$ and $(2)$, we have \[\begin{align} \dfrac{\sin \angle SRB}{\sin \angle BRN} = \dfrac {\sin \angle DRM}{\sin \angle LRD} \tag {3} \end {align*}\] Then, $\angle SRB = \angle DRM$. So $D,R,B$ are colinear. Similarly, $A,R,C$ are colinear. Thus, $R=K$. Stewart for isosceles $\triangle QLN$ gives \[\begin{align} QN^2 - KL \cdot KN = QK^2 \tag {4} \end {align*}\] Stewart for isosceles $\triangle PMS$ gives \[\begin{align} PM^2 - KM \cdot KS = PK^2 \tag {5} \end {align*}\] Since power of $K$ is $KL\cdot KN = KS \cdot KM$, substracting $(5)$ from $(4)$ gives \[\begin{align} QK^2 - PK^2 = QN^2 - PM^2 \tag {6} \end {align*}\] Also $QN^2 = OQ^2 - ON^2$ and $OP^2 = OP^2 - OM^2$. So \[ \begin{align} QK^2 - PK^2 = OQ^2 - OP^2 \tag {7} \end {align*}\] This is equivalent to $PQ \perp OK$. Let $OK \cap PQ = \{H\}$. Since $HPMO$ is cyclic, $\angle OPM = \angle MHO$. We also have $\angle OPM = \angle KMO$. So $\angle OHM = \angle KMO$. This is equivalent to $OK\cdot OH = OM^2$. $\blacksquare$

orl wrote: Quadrilateral $ ABCD$ has an inscribed circle which centered at $ O$ with radius $ r$. $ AB$ intersects $ CD$ at $ P$; $ AD$ intersects $ BC$ at $ Q$ and the diagonals $ AC$ and $ BD$ intersects each other at $ K$. If the distance from $ O$ to the line $ PQ$ is $ k$, prove that $ OK\cdot\ k = r^2$.