Hint: Consider the circumcircle of $ABC$. Clearly $D$ is outside this, so say $E$ is where $BD$ meets this circle ($B\not= E$). Translate the condition $\angle ABD+\angle ACB=\angle ACD+\angle ADB$ into something easier to work with.

mistake by bows and segments changed, so the perpendicularity wasn't proven, but has to be true.

SCP wrote: Also $\angle ACD + \angle BDC = \angle BDA + \angle CAD$ both $90$ degrees I don't understand your solution SCP... can you explain what you mean? Here's how I finished it: $\angle DAE=\angle ECD$ so $\frac{DA}{\sin\angle AED}=\frac{DE}{\sin\angle DAE}=\frac{DE}{\sin\angle ECD}=\frac{DC}{\sin\angle DEC}$ so $\frac{DA}{DC}=\frac{\sin\angle AED}{\sin\angle DEC}=\frac{\sin\angle ACB}{\sin\angle BAC}=\frac{AB}{BC}$. Since $ABCD$ is a circumscribed quadrilateral we also have $AB+CD=BC+DA$. Thus $\frac{DA}{DC}=\frac{BC+DA-CD}{BC}$ which rearranges to $(BC-CD)(DA-CD)=0$. So either $BC=CD$ or $DA=CD$. If $BC=CD$ then $AD=DA$ and so $CA$ is an axis of symmetry for the quadrilateral $ABCD$. Then $B$ is the reflection of $D$ through $AC$ and so $AC$ intersects $BD$ at the midpoint of $BD$. If $DA=CD$ then $AB=BC$ so as before $BD$ is an axis of symmetry for the quadrilateral $ABCD$. Then $A$ is the reflection of $C$ through $BD$ and so $BD$ intersects $AC$ at the midpoint of $AC$.

Drytime wrote: Let $ABCD$ be a convex circumscribed quadrilateral such that $\angle ABC+\angle ADC<180^{\circ}$ and $\angle ABD+\angle ACB=\angle ACD+\angle ADB$. Prove that one of the diagonals of quadrilateral $ABCD$ passes through the other diagonals midpoint. Let incircle be the unit circle which touches sides $AB,BC,CD,DA$ at points $X,Y,Z,T$ and let $M$ and $N$ be the points on line $AB$ such that $A$ lays between $N$ and $B$ and $B$ lays between $A$ and $M$ and $\angle MCB=\angle ABD$ and $\angle ADN=\angle ACD$. Now condition becomes $\angle MCA=\angle BDN$ and we have: \[ \{\begin{array}{ccc}\frac{(ac)}{(cd)}=\frac{(ad)}{(dn)}\\\\ \frac{(mc)}{(ac)}=\frac{(ab)}{(bd)} \\\\ \frac{(mc)}{(ac)}=\frac{(bd)}{(dn)}\end{array}\Longrightarrow (bd)^2(ac)^2=(ab)(bc)(cd)(da) (*)\] Where $(uv)$ means $\frac{u-v}{\bar{u}-\bar{v}}$. On the other hand we have $a=\frac{xt}{x+t},b=\frac{xy}{x+y},c=\frac{yz}{y+z},d=\frac{zt}{z+t}$, so $(*)$ after expanding becomes: \[\frac{(y-t)^2}{yt}(xz-yt)=\frac{(z-x)^2}{zt}(xz-yt)\] So $xz=yt\Longrightarrow XYZT$ is rectangle or $\frac{(y-t)^2}{yt}=\frac{(z-x)^2}{zt}\Longleftrightarrow |y-t|=|z-x| \Longrightarrow XYZT$ is isosceles trapezoid. Now it's obvious.

We are going to show that $ ABCD $ is a kite . Note that if the incentre lies on $ AC $ or $ BD $ then it will lead to a kite , so we may assume that triangle $ AIC $ is not degenerate . Let $ O_1 ~,~ O_2 $ be the circumcentre of $ \Delta ABD $ and $ \Delta BCD $ respectively . From the given condition , we have $ \angle O_1 AI = \angle ACI $ so $ AO_1 $ is a tangent to $ (AIC) $ . Let $ O $ be the centre of $ (AIC) $ , then $ AO $ is a tangent to $ (ABD) $ . Moreover , the condition can be converted to $ \angle CBD + \angle CAB = \angle CDB + \angle CAD $ . We can also conclude that $ CO $ is a tangent to $ (BCD) $ . Note that , from the given condition , $ (ABD) $ and $ (BCD) $ do not coincide . Since $ AO = CO $ , $ O $ is on the radical axis of $ (ABD ) $ and $(BCD) $ , which is $ BD $ . Therefore , $ \frac{AB^2}{AD^2} = \frac{BO}{DO} = \frac{BC^2}{CD^2} $ , or $ \frac{AB}{AD} = \frac{BC}{CD} $ but from $ AD - AB = CD - BC $ , $ AD( 1- \frac{AB}{AD} ) = CD( 1 - \frac{BC}{CD} ) = CD( 1- \frac{AB}{AD} ) $ , we have either $ AD = CD $ or $ 1- \frac{AB}{AD} = 0 $ , but both cases will lead to a kite .

dr_Civot wrote: Drytime wrote: Let $ABCD$ be a convex circumscribed quadrilateral such that $\angle ABC+\angle ADC<180^{\circ}$ and $\angle ABD+\angle ACB=\angle ACD+\angle ADB$. Prove that one of the diagonals of quadrilateral $ABCD$ passes through the other diagonals midpoint. Let incircle be the unit circle which touches sides $AB,BC,CD,DA$ at points $X,Y,Z,T$ and let $M$ and $N$ be the points on line $AB$ such that $A$ lays between $N$ and $B$ and $B$ lays between $A$ and $M$ and $\angle MCB=\angle ABD$ and $\angle ADN=\angle ACD$. Now condition becomes $\angle MCA=\angle BDN$ and we have: \[ \{\begin{array}{ccc}\frac{(ac)}{(cd)}=\frac{(ad)}{(dn)}\\\\ \frac{(mc)}{(ac)}=\frac{(ab)}{(bd)} \\\\ \frac{(mc)}{(ac)}=\frac{(bd)}{(dn)}\end{array}\Longrightarrow (bd)^2(ac)^2=(ab)(bc)(cd)(da) (*)\] Where $(uv)$ means $\frac{u-v}{\bar{u}-\bar{v}}$. On the other hand we have $a=\frac{xt}{x+t},b=\frac{xy}{x+y},c=\frac{yz}{y+z},d=\frac{zt}{z+t}$, so $(*)$ after expanding becomes: \[\frac{(y-t)^2}{yt}(xz-yt)=\frac{(z-x)^2}{zt}(xz-yt)\] So $xz=yt\Longrightarrow XYZT$ is rectangle or $\frac{(y-t)^2}{yt}=\frac{(z-x)^2}{zt}\Longleftrightarrow |y-t|=|z-x| \Longrightarrow XYZT$ is isosceles trapezoid. Now it's obvious. I'm sorry but did you use the condition that the sum of two opposite angle is less than 180?

if $CD=CB$ than $AB=AD$(since $ABCD$ is circumscribed giving $AB+CD=BC+AD$) so $ABCD$ is a kite and $CA$ bisects $BD$ assume WLOG $CD<CB$ the given inequality says that the second intersection of circle $ABD$ and $AC$(point $E$) is outside segment $AC$ the angle equality gives $\angle EBC=\angle ACB-\angle AEB=\angle ACB-\angle ADB=\angle ACD-\angle ABD=\angle ACD-\angle AED=\angle EDC$ so since $EC=EC$ the circumradii of $ECD$ and $ECB$ than we can chose $G$ on circle $ECB$ such that $CG=CD$ and $E,G$ are on the same side of $BC$ now $\angle CBG=\angle DEC=\angle ABD$ and $\angle BGC=\angle BEC=\angle BDA$ so $\triangle BGC\sim\triangle BAD$ and $BA/AD=CB/CG=CB/CD=k>1$ since $ABCD$ is circumscribed $CB-CD=BA-AD$ so $CD*(k-1)=AD*(k-1)$ since $k>1$ than $AD=CD$ and i.e. $AB=BC$ so $ABCD$ is a kite and $DB$ bisects $AC$

Consider the inversion of pole $A$ which fixes $C$, and let $B'$ and $D'$ be the images of $B$ and $D$ under this inversion. We have $\angle AD'B'+\angle AB'C=\angle AB'D'+\angle AD'C\Rightarrow \angle CB'D'=\angle CD'B'$, hence $CB'=CD'$. But $B'C/AC=BC/AB$ and $D'C/AC=DC/AD$, so it follows that $AB\cdot CD=AD\cdot BC$. Using the fact that $AB+CD=AD+BC$, we easily get that $ABCD$ is either a kite or a losange.

Let $P$ and $P^\prime$ be points outside $ABCD$ such that $\triangle{APB}\sim \triangle{BAC}$ and $\triangle{AP^\prime D}\sim\triangle{DCA}$. Let $\{K\}=BP\cap DP^\prime$ and observe that $KB$ is tangent to $\triangle{ABC}$ and $KD$ si tangent to $\triangle{ADC}$. As $KB=KD$, from the power of point we get that $K$ is on the radical axe of the two circles, namely $AC$. From angle chasing, we get that $CA$ is tangent to the circumcircles of $\triangle{PAB}$ and $\triangle{P^\prime D A}$.So $KP\cdot KB=KA^2=KP^\prime\cdot KD$, but as $\triangle{KBD}$ is isosceles, we'll get $PB=P^\prime D$. Due to the similarities between triangle, $PB=\dfrac{AC\cdot AB}{BC},\ P^\prime D=\dfrac{AC\cdot AD}{CD}$. Therefore, we have $AD\cdot BC= AB \cdot CD$, and the conclusion follows.

Let $M$ be on $BD$ such that $ABCM$ is cyclic.Now,we get $<MAD=<MCD$ and now appl, sine rule in $AMD,CMD$ and $ABC$(this can easy be avoided by proving similarities) to get $AB/BC=CD/AD$ and now it is trivial cause $AB+CD=BC+AD$,so we are finished.

I have found another solution. Just take the tangents through$ B$ and $D$ WRT the circumcircles of $ABC$ and $ADC$. Let them meet at a point $E$. From the angle condition we deduce that $BE=DE$. $E$ has the same power WRT the considered circumcircles,so $E$ lies on $AC$(their radical axis). Now apply the law of sines in triangles $BCE$ and $DCE$: $\frac {BE}{\sin\angle BCA}=\frac {CE}{\sin\angle CBE} $ i.e. $\frac {BE}{CE}=\frac {\sin\angle BCA}{\sin\angle BAC}=\frac {AB}{BC}$. Doing the same in the other part, $\frac {DE}{CE}=\frac {\sin\angle ACD}{\sin\angle DAC}=\frac {AD}{DC}$. We deduce that $\frac {AB}{BC}=\frac {AD}{DC} $,and using that $AB+CD=AD+BC$, after some computations we are done .

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