bilarev wrote: In the quadrangle $ABCD$ ($\angle BAD+\angle ADC>\pi$) is inscribed a circle with center $I.$ A line throught $I$ intersects $AB$ and $CD$ in points $X$ and $Y$ respectively such that $IX=IY.$ Prove that $AX.DY=BX.CY.$ bilarev, i think that your problem, in this form, is wrong! Your problem implies that M, N and I are in the same line, where M and N are the points where the incircle touchs AB and CD! (The congruence of the triangles MXI and NYI) EDIT: I'm Wrong! See my solution in my next post!

e.lopes wrote: bilarev, i think that your problem, in this form, is wrong! Your problem implies that M, N and I are in the same line, where M and N are the points where the incircle touchs AB and CD! (The congruence of the triangles MXI and NYI) You are wrong...think a little bit about the positin of the points...(M and N may lie on segments AX and DY).

bilarev wrote: e.lopes wrote: bilarev, i think that your problem, in this form, is wrong! Your problem implies that M, N and I are in the same line, where M and N are the points where the incircle touchs AB and CD! (The congruence of the triangles MXI and NYI) You are wrong...think a little bit about the positin of the points...(M and N may lie on segments AX and DY). Thank you for your correction. (i wasn't using that A+D > 180) Here is my solution, with the points in your correct places! _ Let M and N the points where the incircle touchs AB and DC! We know that the triangles MXI and NYI are congruents! So, $<MIX = <NIY = 90-\frac{MIN}{2}= \frac{B+C}{2}$. Let $MX = NY = x$. First, we know that $x = \frac{r}{tan(\frac{B+C}{2})}$. See that we have to prove that $(AM+x)(DN+x) = (MB-x)(NC-x)$. This is the same that prove that $\frac{r}{tan(\frac{B+C}{2})}= \frac{MB.NC-AM.DN}{MB+MA+NC+ND}$. We easily see that $r(MB+MA+NC+ND) = [ABCD]$. So, we have to prove that $[ABCD] = (MB.NC-AM.DN)tg(\frac{B+C}{2})$. We easily see that $[ADI]+[BCI] = [ABI]+[CDI]$. So, $[ABCD] = 2([ADI]+[BCI]) = (AI.DI+BI.CI)sen(\frac{B+C}{2})$. So, we have to prove that $BM.CN-DN.AM = (AI.DI+BI.CI)cos(\frac{B+C}{2})$. By Cosines Law in the triangles AID and BIC: $AI.DIcos(\frac{B+C}{2})+BI.CIcos(\frac{B+C}{2}) = \frac{(AI^{2}+DI^{2}-AD^{2})-(BI^{2}+CI^{2}-BC^{2})}{2}$ $= \frac{(r^{2}+AM^{2}+r^{2}+DN^{2}-(AM+DN)^{2})-(r^{2}+BM^{2}+r^{2}+NC^{2}-(BM+NC)^{2})}{2}$ $= BM.CN-DN.AM$!

Let $\rho$ be radius of (I). Let P, Q be tangency points of AB, CD with (I). Let AB, CD meet at E, on the opposite side of DA than I (because $\angle D+\angle A > \pi$). EI bisects $\angle DEA \equiv \angle CEB,$ $XY \perp EI.$ Reflect E in XY into E', EXE'Y is a rhombus, E'X, E'Y are tangents of (I) at P', Q', and E'P' = E'Q' = EP = EQ. Let DA meet E'X at F and E'Y at G. (I) is the E-excircle of the $\triangle DEA,$ F-excircle of the $\triangle FXA,$ and G-excircle of the $\triangle DYG$ and these 3 triangles are similar, having equal angles. Let $r_{d},\ r_{e}= \rho,\ r_{a}$ be exradii, r inradius, $s = EP = EQ$ semiperimeter, and $S = \sqrt{r r_{d}r_{e}r_{a}}$ area of the $\triangle DEA$ (Heron's formula for S). Then $\frac{AX}{EA}= \frac{r_{e}}{r_{d}},$ $\frac{DY}{ED}= \frac{r_{e}}{r_{a}},$ $AX \cdot DY = EA \cdot ED \cdot \frac{r_{e}^{2}}{r_{d}r_{a}}= \frac{2S}{\sin E}\cdot \frac{r_{e}^{3}r}{S^{2}}= \frac{2 r_{e}^{3}}{s \sin E}= \frac{2 \rho^{3}}{EP \sin E}$ Similarly, $BX \cdot CY = \frac{2 \rho^{3}}{E'P' \sin E'}= \frac{2 \rho^{3}}{EP \sin E},$ hence $AX \cdot DY = BX \cdot CY.$

Sorry, this problem is equivalently with a nice and easy old problem : Quote: An equivalent enunciation. Let $C(I)$ be the incircle of the triangle $ABC$. The points $E\in AC$ and $F\in AB$ belong to the circle $w$. Denote the points $X\in (AC)$ and $Y\in (AB)$ for which $I\in XY$ and $IX=IY$. Prove that for any points $M\in (AE)$ and $N\in (AF)$ for which the line $MN$ is tangent to the circle $w$, the product $XM\cdot YN$ is constant, more precisely, $XM\cdot YN=IX^{2}$. Proof. Denote $\{\begin{array}{c}m(\widehat{BNM})=2y\\\ m(\widehat{CMN}=2x\end{array}$. Observe that $\boxed{x+y=90^{\circ}+\frac{A}{2}}$ and $\{\begin{array}{c}m(\widehat{XIM})=90^{\circ}+\frac{A}{2}-x\ ,\ m(\widehat{XIC})=\frac{B}{2}\\\ m(\widehat{YIN})=90^{\circ}+\frac{A}{2}-y\ ,\ m(\widehat{YIB})=\frac{C}{2}\end{array}$. Therefore, $\{\begin{array}{c}\frac{XM}{XC}=\frac{IM}{IC}\cdot\frac{\sin\widehat{XIM}}{\sin\widehat{XIC}}=\frac{\sin\frac{C}{2}}{\sin x}\cdot\frac{\cos(x-\frac{A}{2})}{\sin\frac{B}{2}}\\\\ \frac{YN}{YB}=\frac{IN}{IB}\cdot\frac{\sin\widehat{YIN}}{\sin\widehat{YIB}}=\frac{\sin\frac{B}{2}}{\sin y}\cdot\frac{\cos(y-\frac{A}{2})}{\sin\frac{C}{2}}\end{array}$ $\implies$ $\frac{XM}{XC}\cdot\frac{YN}{YB}=$ $\frac{\cos(x-\frac{A}{2})\cos(y-\frac{A}{2})}{\sin x\sin y}=$ $\frac{\cos (x-y)+\cos (x+y-A)}{\cos (x-y)-\cos (x+y)}=$ $\frac{\cos (x-y)+\cos(90^{\circ}-\frac{A}{2})}{\cos (x-y)-\cos(90^{\circ}+\frac{A}{2})}=$ $\frac{\cos (x-y)+\sin\frac{A}{2}}{\cos (x-y)+\sin\frac{A}{2}}=1$ $\implies$ $XM\cdot YN=XC\cdot YB$ (constant). Observe that the constant $k\equiv XC\cdot YB=XA\cdot XE=IX^{2}=$ $(IA\cdot\tan\frac{A}{2})^{2}=$ $\frac{bc(p-a)}{p}\cdot\frac{(p-b)(p-c)}{p(p-a)}=\frac{bc(p-b)(p-c)}{p^{2}}$. Remarks. $1\blacktriangleright$ The points $X$, $Y$ belong to the circle which is tangent to the rays $[AB$, $[AC$ and which is internal tangent to the circumcircle of $\triangle ABC$. $2\blacktriangleright XA=YA=\frac{bc}{p}$. Memorize the following relations $\{\begin{array}{c}p(p-a)+(p-b)(p-c)=bc\\\ p(p-a)-(p-b)(p-c)=bc\cdot\cos A\end{array}$, where $a+b+c=2p$. $3\blacktriangleright$ From the above identities you can prove very easily the following remarkable formulas : $\{\begin{array}{c}\sin\frac{A}{2}=\sqrt{\frac{(p-b)(p-c)}{bc}}\\\\ \cos\frac{A}{2}=\sqrt{\frac{p(p-a)}{bc}}\\\\ \tan\frac{A}{2}=\sqrt{\frac{(p-b)(p-c)}{p(p-a)}}\end{array}$

It is easy to show by angle chasing (but I am too lazy to write it), that $ \angle AIX = \angle IDY$ and $ \angle BIX = \angle ICY.$ So, from the similar triangles $ \bigtriangleup AIX$ and $ \bigtriangleup IDY,$ we conclude that $ \frac {AX}{IY} = \frac {IX}{DY}$ $ \Longrightarrow$ $ (AX)\cdot (DY) = (IX)^{2}$ $ ,(1)$ Also, from the similar triangles $ \bigtriangleup BIX$ and $ \bigtriangleup ICY,$ we conclude that $ \frac {BX}{IY} = \frac {IX}{CY}$ $ \Longrightarrow$ $ (BX)\cdot (CY) = (IX)^{2}$ $ ,(2)$ Hence, from $ (1),$ $ (2)$ $ \Longrightarrow$ $ (AX)\cdot (DY) = (BX)\cdot (CY)$ and the proof is completed. Kostas Vittas.

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In case one is being too stupid to notice the similar triangles, this problem can also be bashed easily. Let the lengths of the tangents to the incircle from vertices $A, B, C, D$ be $a, b, c, d$, respectively. Let $AB,CD$ touch the incircle at $M,N$, respectively. Lemma: If $r$ is the inradius of tangential quadrilateral $ABCD$ then \[r^2 = \frac {abc + abd + acd + bcd}{a + b + c + d}.\] Proof: We can easily see that $A + B + C + D = 2\pi$, $\tan(A/2) = r/a$, $\tan(B/2) = r/b$, $\tan(C/2) = r/c$, and $\tan(D/2) = r/d$. Then \[0 = \tan\pi = \tan\left(\frac{A}{2}+\frac{B}{2}+\frac{C}{2}+\frac{D}{2}\right)\\ = \frac {\sum\tan\frac{A}{2} - \sum\tan\frac{A}{2}\tan\frac{B}{2}\tan\frac{C}{2}}{1 - \sum\tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{A}{2}\tan\frac{B}{2}\tan\frac{C}{2}\tan\frac{D}{2}},\]so clearing the denominator, \[\frac{r}{a} + \frac{r}{b} + \frac{r}{c} + \frac{r}{d} = \frac{r^3}{abc} + \frac{r^3}{abd} + \frac{r^3}{acd} + \frac{r^3}{bcd}.\]and the lemma follows.$\blacksquare$ Now let $MX=NY=e$. We need to show \[AX\cdot DY=BX\cdot CY \Longleftrightarrow (a+e)(d+e)=(b-e)(c-e) \Longleftrightarrow e=\frac{bc-ad}{a+b+c+d}.\]Note that \begin{align*} \frac{e}{r}=\tan\angle{MIX}=\tan\frac{A+D-\pi}{2}=-\frac{1}{\tan\frac{A+D}{2}}&=\frac{\tan\frac{A}{2}\tan\frac{D}{2}-1}{\tan\frac{A}{2}+\tan\frac{D}{2}}\\ &=\frac{\frac{r}{a}\cdot\frac{r}{d}-1}{\frac{r}{a}+\frac{r}{d}}=\frac{r^2-ad}{r(a+d)}. \end{align*}Thus by the lemma, \[e=\frac{r^2-ad}{a+d}=\frac{\frac{abc+abd+acd+bcd}{a+b+c+d}-ad}{a+d}=\frac{abc+bcd-a^2d-ad^2}{(a+b+c+d)(a+d)}=\frac{bc-ad}{a+b+c+d},\]as desired.