Let $T$ be the second intersection of the circle $(COD)$ with $AB$ ($O$ is the center of the circle which touches $BC,CD,DA$). We have $\angle DTA=\angle DCO=\frac{\angle DCB}2=\frac\pi 2-\frac{\angle DAB}2$, which means that $AD=AT$. In the same way we prove that $BC=BT$, and the conclusion follows.

grobber wrote: Let $T$ be the second intersection of the circle $(COD)$ with $AB$ ($O$ is the center of the circle which touches $BC,CD,DA$). We have $\angle DTA=\angle DCO=\frac{\angle DCB}2=\frac\pi 2-\frac{\angle DAB}2$, which means that $AD=AT$. In the same way we prove that $BC=BT$, and the conclusion follows. Can you explain what do you mean by circle COD is it the circle around the triangle DOC,or the circle centared at O

Trigonometrically : Let the tangency points be $ M, N, P$ ( on $AD, DC, CB$) , and $\angle OAD=x, \angle OBC=y$. After some computations, using the sinus theorem, we get that the conclusion is equivalent with $ \cot (x) + \cot (y) + \tan (\frac{x}{2}) + \tan(\frac{y}{2}) = \frac{1}{\sin(x)}+\frac{1}{\sin(y)}$ . But $ \tan (a)+\tan (b) = \frac{\sin(2a)}{1+\cos(2a)}+\frac{\sin(2b)}{1+\cos(2b)}$, so the conclusion holds.

Teodor95 wrote: Can you explain what do you mean by circle COD is it the circle around the triangle DOC,or the circle centared at O $(COD)$ is obviously a circumcircle of the triangle $COD$

I have a different version of the problem where ABCD -> EDCB. [asy][asy]defaultpen(linewidth(0.8)+fontsize(10));size(6cm); dotfactor = 4.5; real y = 25, x = 105, z = 170; pair O = origin, Y = dir(y), X = dir(x), Z = dir(z), M = extension(O, O+dir(x-90), Y, Y+dir(y-90)), N = extension(O, O+dir(x+90), Z, Z+dir(z+90)), B = extension(Z, Z+dir(z-90), X, X+dir(x+90)), C = extension(X,X+dir(x+90), Y, Y+dir(y-90)), D = reflect(O,O+dir((y+z)/2))*N, E = reflect(O,O+dir((y+z)/2))*M; markscalefactor = 0.015; draw(B--E--D--C--B--O--C^^rightanglemark(O,X,C)^^rightanglemark(O,Y,C)^^rightanglemark(O,Z,B)^^Arc(O,1,x+90, x+90-y-z)^^E--N); draw(M--N, linetype("2 2")); draw(X--O--Y^^Z--O, linetype("4 4")); pair point = origin+(0,0.01); pair[] p={B,C,D,E,X,Y,Z,M,N,O}; string s = "B,C,D,E,X,Y,Z,M,N,O"; int size = p.length; real[] d; real[] mult; for(int i = 0; i<size; ++i) { d[i] = 0; mult[i] = 1;} string[] k= split(s,","); for(int i = 0;i<p.length;++i) { dot("$"+k[i]+"$",p[i],mult[i]*dir(point--p[i])*dir(d[i])); } [/asy][/asy] Let $O$ be the center of the circle, and $X$, $Y$, and $Z$ be the tangency points as shown. Construct points $M$ and $N$ on lines $CD$ and $BE$ respectively so that $MN$ is parallel to $BC$ and passes through $O$. Notice that $\angle BNM = 180 - \angle NBC = \angle EDC$ and $\angle CMN = 180 - \angle BCD = \angle BED$. It follows that $\triangle OZE \cong OYM$ and consequently $\triangle ONE \cong \triangle ODM$. Furthermore, $\angle NBO = \angle CBO = \angle NOB$, so $NO = NB$ and $MO = MC$ similarly. Thus, we have that $BE + CD = BN - EN + CM + MD = BN + CM = NO + MO = OD + OE = DE$.

I have a solution similar to dragon96. But it is also visual expression of drEdrE's trigonometric solution. Let the circle touch $BC$ at $P$, $AD$ at $Q$. Let $O$ be its center. Let $\angle BAD = 2\alpha$. $\angle DCB = 180^\circ - 2\alpha \Rightarrow \angle OCP = 90^\circ - \alpha$. Let $S$ be a point on $[QD$ such that $PC=QS$. Also, $QO=OP$ and $\angle SQO=\angle CPO = 90^\circ$ yield $\triangle CPO \cong \triangle SQO$, so, $\angle QSO = \angle PCO = 90^\circ - \alpha$. In $\triangle ASO$, since $\angle SOA = 180^\circ - 2\alpha - (90^\circ - \alpha) = 90^\circ - \alpha = \angle ASO$, we get $AO=AS=AQ+CP$. Similarly, $BO=BP+QD$. $AB=AO+OB=AQ+CP+BP+QD=AD+BC$. $\blacksquare$

In grobber's solution, I do not see why $\angle DTA=\angle DCO$

Mudkipswims42 wrote: It is clear that $OD$ bisects $\angle CDA$ and $OC$ bisects $\angle DCB$ (because of the tangents we can quickly form 2 congruent triangles) Also, if we let $E=\overline{BC}\cap\overline{AD}$ we can see that $O$ is the $E$-excenter of $\triangle CDE$ , from which angle bisection follows clearly from the excircle definition.

Let $O$ be the center of the circle that has a diameter of $AB$ and let $E,F,G$ be the perpendiculars from $O$ to the tangency points on lines $AD, DC, CB$ respectively. Let $D'$ be on $AB$ such that $AD=AD'$. It suffices to prove that $D'B=BC$. Claim 1: $DOD'C$ is cyclic. Proof: From $AD=AD'$, we have $\angle ADD'=\angle AD'D$ so $\angle DAD'=180^{\circ}-2\angle AD'D.$ By $HL,$ we have $\triangle OFC\cong \triangle OGC$ so $\angle OCF=\angle OCG$. Due to cyclic quadrilateral $ABCD$, $$2\angle DCO=\angle DCB=180^{\circ}-\angle DAB=2\angle AD'D,$$proving the quadrilateral to be cyclic.$\Box$ Claim 2: $\angle CD'B=90^{\circ}-\frac{1}{2}\angle B$. Proof: Note that $\angle FOE=180^{\circ}-\angle EDF=\angle B.$ Because $\triangle OED\cong \triangle OFD$, $\angle DOF=\angle DOE=\frac{1}{2}\angle B$. By Claim 1, $$\angle CD'B=\angle CDO=90^{\circ}-\angle DOF=\angle CD'B=90^{\circ}-\frac{1}{2}\angle B.$$$\Box$ By Claim 2, we have $$\angle D'CB=180^{\circ}-\angle CD'B-\angle CBD'=180^{\circ}-90^{\circ}+\frac{1}{2}\angle B-\angle B=90^{\circ}-\frac{1}{2}\angle B.$$Thus, $D'B=BC$ and we are done. $\blacksquare$

Sorry for again reviving an ancient thread, but is there a way to prove the converse of this statement?

anyone? anyone?

Booper wrote: Sorry for again reviving an ancient thread, but is there a way to prove the converse of this statement? For the converse, we will prove that given $AB=BC+AD$, there exists a circle tangent to the three sides having centre on $AB$ Let $T$ be a point on $AB$ such that $AT=AD$ and $TB=BC$ then let the circumcircle of $TDC$ intersect $AB$ at point $O$. If $O \neq T$: then, it is easy to see that $\angle DTC = \angle DOC = \frac{\angle A + \angle B}{2}$ since, $\angle ATD = \frac{\angle C}{2} \Rightarrow \angle OTC = \frac{\angle D}{2}$ $$\Rightarrow \angle ODC = \angle OTC = \frac{\angle D}{2}$$Hence, $OD$ is angle bisector of $\angle ADC$ similarly $OC$ bisects $\angle BCD$. hence point $O$ is equidistant from $AD,BC,CD$ if $O$ and $T$ coincide: then, $AB$ is tangent to $(DOC)$ hence, $\angle OCD=\angle AOD = \frac{\angle C}{2}$ again, $OC$ bisects $\angle BCD$ and we are done!

Draw the perpendiculars from $O$--the center of the circle tangent to the three sides of the circle whose center lies on $AD$--to $AD,DC$, and $CB$. Call their intersections $X,Y,Z$. Then, if $\angle OAD = 2a$ and $\angle OBC =2a$, $\angle XOD = \angle DOY=a$ and $\angle YOC = \angle COZ = b$. Let the radius of circle $O$ be $1$. We have that $$AD+BC = \frac{1}{\tan(2b)} + \tan(b) + \frac{1}{\tan(2a)} + \tan(a) = \frac{1-\tan^2(b)}{2\tan(b)} + \frac{1-\tan^2(a)}{2\tan(a)}+\tan(a) + \tan(b) =\frac{1+\tan^2(b)}{2\tan(b)} +\frac{1+\tan^2(a)}{2\tan(a)} $$$$\frac{1+\tan^2(b)}{2\tan(b)} = \frac{1+\frac{\sin^2(b)}{\cos^2(b)}}{2\frac{\sin(b)}{\cos(b)}} = \frac{1}{2\sin(b)\cos(b)} = \frac{1}{\sin(2b)}$$Therefore, $AD+BC = \frac{1}{\sin (2b)} + \frac{1}{\sin (2a)} = AB$.

Let the circle, call it $\omega$, have radius $r$ and center $O$. Let $BC$ be tangent to $\omega$ at $X$, $CD$ at $Y$, and $DA$ at $Z$. Then we have \begin{align*} AB&=AO+OB=\frac{r}{\sin A}+\frac{r}{\sin B}\\ AD&=AZ+ZD=r\tan (90^\circ-A)+r\tan B/2\\ BC&=BX+XC=r\tan (90^\circ-B)+r\tan A/2. \end{align*}So we need to show that \[\frac{1}{\sin A}=\tan (90^\circ-A)+\tan A/2=\frac{\cos A}{\sin A}+\frac{\sin A/2}{\cos A/2}.\]Let $\sin A/2=x,\cos A/2=y$ with $x^2+y^2=1$ so we need to show \[\frac{1}{2xy}=\frac{2y^2-1}{2xy}+\frac{x}{y}\iff 1=2y^2-1+2x^2\]which is trivial.

Let the tangent circle be $\omega$, the center of $\omega$ be $I_p$, $\omega$ touch $BC, CD, DA$ at $Z, Y, X$ respectively, and $\overrightarrow{AD} \cap \overrightarrow{BC} = P$. Then, $\omega$ is the $P$-excircle of $PCD$. Now, let the foot of the $P$-altitude to $CD$ be $E$. Claim: $PEC \sim I_pXA$ and $PED \sim I_pZB$. Proof. Notice $\angle I_pAX = \angle BAD = \angle PCD = \angle PCE$ so $PEC \sim I_pXA$ by AA. An analogous process yields $PED \sim I_pZB$ as desired. $\square$ Denote the radius of $\omega$ as $R$. It's well-known $DX = DY$ and $CY = CZ$. Then, $$AD + BC = AX + XD + BZ + ZC = \frac{R}{PE} \cdot EC + DY + \frac{R}{PE} \cdot ED + CY$$$$= \frac{R}{PE} \left(EC + ED \right) + CD = CD + \frac{R}{PE} \cdot CD = \frac{PE + R}{PE} \cdot CD$$so it suffices to show $\frac{PE + R}{PE} = \frac{AB}{CD}$, which is equivalent to proving $PE + R$ has the same length as the altitude from $P$ to $AB$. Let the foot of the $P$-altitude to $AB$ be $F$. Now, reflect $PCD$ in the angle bisector of $\angle APB$. Obviously, $C' \in PA$ such that $PC' = PC$ and $D' \in PB$ such that $PD' = PD$. Claim: $C'D' \parallel AB$. Proof. By POP, $$PD \cdot PA = PC \cdot PB \implies \frac{PC}{PA} = \frac{PD}{PB} = \frac{PC'}{PA} = \frac{PD'}{PB}.$$$\square$ Thus, $PE + R = PE + YI_p = PE' + Y'I_p = PF$ and we're done. $\blacksquare$ Remark: My inspiration for reflecting $PCD$ was thinking about isogonality (specifically $O$ and $H$) and how I could obtain the altitude from $P$ to $AB$. Also, $I_p$ is the unique point that allows use to use this reflection since $I_p \in AB$ and $I_p \in PI$, so $I_p' = I_p$.

Mehhhhh nice problem! Let $P$ be a point on side AB such that $BC=BP$, we just need to prove $AD=AP$. Hence let $OCB=x$ and $OCP=y$. First note $DCB=2x$ (because of the tangency, $OC$ bissects $DCB$), and then $DAB=180-2x$ (because of $ABCD$ cyclic). Then note we just need $DPA=x$. For this $DCOP$ has to be cyclic. Since $CBA=180-2x-2y$, we have $ADC=2x+2y$. And since $OD$ bissects $ADC$, we have $ODC=x+y$. AS $BC=BP$, we have $CPO=x+y=CDO$, and we finish the problem, as we prove $DCOP$ is cyclic.

[asy][asy] import olympiad; size(10cm); pair A, B, C, D, E, F, G, O; A = (0, 0); B = (9, 0); C = (8.52, 3.94); draw(circumcircle(A, B, C)); D = (2.21, 5.59); draw(A--D); draw(B--C); draw(A--B); draw(C--D); O = (4.7, 0); E = (0.63, 1.61); F = (8.94, 0.51); draw(O--D); draw(O--C); draw(O--E); draw(O--F); G = (5.91, 4.62); draw(G--O); draw(rightanglemark(O, F, C)^^rightanglemark(D, E, O)^^rightanglemark(O, G, C)); dot(A^^B^^C^^D^^E^^F^^G^^O); label("$A$", A, dir(180)); label("$B$", B, dir(0)); label("$C$", C, dir(45)); label("$D$", D, dir(90)); label("$E$", E, dir(135)); label("$F$", F, dir(45)); label("$G$", G, dir(90)); label("$O$", O, dir(270)); [/asy][/asy] $O$ is the center of the circle that is tangent to sides $AD$, $DC$, and $BC$. Since $O$ is the center of that circle, we know that $OE=OG=OF$. Furthermore, since the three sides are tangent to the circle, $\angle{DEO}=\angle{CGO}=\angle{CFO}=90^\circ$. Claim 1. $\sin(x)\sin(2x-90)+\cos(x)\cos(2x-90)=\sin(x)$. Proof. Because \begin{align*} \sin(\alpha)\sin(\beta)&=\frac{1}{2}(\cos(\alpha-\beta)-\cos(\alpha+\beta)) \\ \cos(\alpha)\cos(\beta)&=\frac{1}{2}(\cos(\alpha+\beta)+\cos(\alpha-\beta)) \end{align*}we have \begin{align*} \sin(x)\sin(2x-90)+\cos(x)\cos(2x-90)&=\frac{1}{2}(\cos(x-90)-\cos(3x-90))+\frac{1}{2}(\cos(3x-90)+\cos(x-90)) \\ &= \cos(x-90) \\ &= \sin(x) \quad \blacksquare \end{align*} Now let $OE=OG=OF=x$ and let $\angle{EDO}=\angle{GDO}=\alpha$ and let $\angle{GCO}=\angle{CFO}=\beta$. Now we have \begin{align*} AD&=x\left(\tan(2\beta-90)+\frac{1}{\tan(\alpha)}\right) \\ BC &= x\left(\tan(2\alpha-90)+\frac{1}{\tan(\beta)}\right) \\ AB &= x\left(\frac{1}{\cos(2\beta-90)}+\frac{1}{\cos(2\alpha-90)}\right) \end{align*}So it remains to prove \[ \tan(2\beta-90)+\frac{1}{\tan(\beta)}+\tan(2\alpha-90)+\frac{1}{\tan(\alpha)}=\frac{1}{\cos(2\beta-90)}+\frac{1}{\cos(2\alpha-90)} \]Because $\tan(x)=\frac{\sin(x)}{\cos(x)}$ we have \[ \frac{\sin(2\beta-90)}{\cos(2\beta-90)}+\frac{\cos(\beta)}{\sin(\beta)}+\frac{\sin(2\alpha-90)}{\cos(2\alpha-90)}+\frac{\cos(\alpha)}{\sin(\alpha)}=\frac{1}{\cos(2\beta-90)}+\frac{1}{\cos(2\alpha-90)} \]So if we prove that \[ \frac{\sin(2x-90)}{\cos(2x-90)}+\frac{\cos(x)}{\sin(x)}=\frac{1}{\cos(2x-90)} \]we would be done. Now notice \begin{align*} \frac{\sin(2x-90)}{\cos(2x-90)}+\frac{\cos(x)}{\sin(x)} &= \frac{\sin(2x-90)\sin(x)+\cos(x)\cos(2x-90)}{\sin(x)\cos(2x-90)}=\frac{1}{\cos(2x-90)} \\ &\implies \sin(2x-90)\sin(x)+\cos(x)\cos(2x-90)=\sin(x) \end{align*}which is true because of Claim 1. $\blacksquare$

Let $T$ be the point such that $DA = AT$. Claim: $T$ lies on $(DOC)$. Proof. Because \[ \angle DCO = \frac{1}{2} \angle DCB = \frac{1}{2}(180^{\circ} - \angle BAD) = 90^{\circ} - \frac{1}{2} \angle TAD = \angle DTA. \]$\blacksquare$ [asy][asy] pair X = dir(105); pair Z = dir(160); pair Y = dir(35); pair D = 2*X*Z/(X+Z); pair A = 2/(Z+1/Z); pair B = 2/(Y+1/Y); pair C = 2*X*Y/(X+Y); filldraw(A--B--C--D--cycle, invisible, blue); draw(circumcircle(A, B, C), grey); pair O = origin; draw(arc(O, 1, 0, 180), blue); pair T = -O+2*foot(circumcenter(D, O, C), A, B); filldraw(circumcircle(D, O, C), invisible, deepgreen); draw(D--T--C, deepgreen); draw(D--O--C, lightcyan); dot("$D$", D, dir(D)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$O$", O, dir(270)); dot("$T$", T, dir(270)); /* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ X := dir 105 Z := dir 160 Y := dir 35 D = 2*X*Z/(X+Z) A = 2/(Z+1/Z) B = 2/(Y+1/Y) C = 2*X*Y/(X+Y) A--B--C--D--cycle / 0.1 lightcyan / blue circumcircle A B C / grey O 270 = origin arc O 1 0 180 / blue T 270 = -O+2*foot (circumcenter D O C) A B circumcircle D O C / 0.1 yellow / deepgreen D--T--C / deepgreen D--O--C / lightcyan */ [/asy][/asy] Reversing the previous proof on the other side gives $BC = BT$. So $AB = AT + TB = AD + BC$.

v_Enhance wrote: Let $T$ be the point such that $DA = DT$. Claim: $T$ lies on $(DOC)$. Proof. Because \[ \angle DCO = \frac{1}{2} \angle DCB = \frac{1}{2}(180^{\circ} - \angle BAD) = 90^{\circ} - \frac{1}{2} \angle TAD = \angle DTA. \]$\blacksquare$ [asy][asy] pair X = dir(105); pair Z = dir(160); pair Y = dir(35); pair D = 2*X*Z/(X+Z); pair A = 2/(Z+1/Z); pair B = 2/(Y+1/Y); pair C = 2*X*Y/(X+Y); filldraw(A--B--C--D--cycle, invisible, blue); draw(circumcircle(A, B, C), grey); pair O = origin; draw(arc(O, 1, 0, 180), blue); pair T = -O+2*foot(circumcenter(D, O, C), A, B); filldraw(circumcircle(D, O, C), invisible, deepgreen); draw(D--T--C, deepgreen); draw(D--O--C, lightcyan); dot("$D$", D, dir(D)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$O$", O, dir(270)); dot("$T$", T, dir(270)); /* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ X := dir 105 Z := dir 160 Y := dir 35 D = 2*X*Z/(X+Z) A = 2/(Z+1/Z) B = 2/(Y+1/Y) C = 2*X*Y/(X+Y) A--B--C--D--cycle / 0.1 lightcyan / blue circumcircle A B C / grey O 270 = origin arc O 1 0 180 / blue T 270 = -O+2*foot (circumcenter D O C) A B circumcircle D O C / 0.1 yellow / deepgreen D--T--C / deepgreen D--O--C / lightcyan */ [/asy][/asy] Reversing the previous proof on the other side gives $BC = BT$. So $AB = AT + TB = AD + BC$. İ think there is a typo it should be DA=AT

Let $O\in AB$ be the center of circle and $\odot (OCD)$ meets $AB$ again at $E$. Since $$\angle BEC=\angle CDO=\frac{\angle ADC}{2}=\frac{\pi -\angle CBE}{2}$$we obtain $|BC|=|BE|$. Analogously $|AD|=|AE|,$ so the conclusion follows.

Let $O$ be the center of the circle and let $E$ be the point on $|AB|$ such that $|AD|=|AE|$, it suffices to show $|BC|=|BE|$. Claim: $DEOC$ is cyclic. Proof: Note that $$\angle{DCO}=\frac{1}{2}\angle{DCO}=\frac{1}{2}(180-\angle{EAD})=\angle{DEA}.$$ Claim: $|BC|=|BE|$. Proof: Note that $$\angle{BEC}=\angle{ODC}=\frac{1}{2}\angle{ADC}=\frac{1}{2}(180-\angle{CBA})\Rightarrow$$$$\Rightarrow\angle{ECB}=180-\angle{CBA}-\frac{1}{2}(180-\angle{CBA})=\frac{1}{2}(180-\angle{CBA})=\angle{BEC}.$$ From last claim we get $|AB|=|AE|+|EB|=|AD|+|BC|$, as wanted.

Let $T$ be on segment $AB$ such $AT = AD$. Then, since $2\measuredangle DTO = 2\measuredangle DTA = \measuredangle DAB$ and $2\measuredangle DCO = \measuredangle DCB$, it follows that $DTOC$ is also cyclic. As such, \[ 2\measuredangle BTC = 2\measuredangle OTC = 2\measuredangle ODC = \measuredangle ADC = \measuredangle ABC \]implies that $\measuredangle BTC = \measuredangle TCB$ and $BT = BC$. Thus, the result holds.

egmo chosen problems are very low mohs on purpose Let $E$ be on $AB$ such that $AD=AE$. We have $\angle DEO=180^{\circ}-\angle DEA=180^{\circ}-\frac{180^{\circ}-\angle DAB}{2}=180^{\circ}-\frac{\angle DCB}2=180^{\circ}-\angle DCO\implies CDEO$ is cyclic, so $\angle CEB=\angle CDO=\frac{180^{\circ}-\angle CBE}2\implies BC=BE$. Finally, $AD+BC=AE+BE=AB$, as desired. $\blacksquare$ Note: It can easily be checked that this holds analogously no matter whatever side of O E is on, in this case E was on the left side of O.

Let $T$ be in $\overline{AB}$ so that $AT = DT.$ From isosceles triangle $\triangle TAD,$ $\angle ATD = \frac{180^\circ - \angle BAD}{2} = \frac{180^\circ - \angle TAD}{2}.$ Since $ABCD$ is cyclic and common tangents form congruent angles, $$\frac{180^\circ - \angle TAD}{2} = \frac{\angle DCB}{2} = \angle DCO.$$This proves that $DTOC$ is cyclic. Note that $\angle ABC = 180^\circ - \angle ADC.$ Now, it’s just straightforward angle chasing: $$\angle BTC = \angle OTC = \angle ODC = \frac{\angle ADC}{2}.$$Hence, $\triangle BTC$ is isosceles and thus $BT = BC.$ Therefore, $AD + BC = AT + BT = AB.$ $\blacksquare$

Construct $X$ on $AD$ such that $AB = AX$ and $CD = CX$, and let $Y = AD \cap (BCX)$, where $Y \neq X$ unless $AD$ is tangent to $(BCX)$. Noting the cyclicity and the isosceles triangles, we have \[\angle CBY = \angle CXD = \frac{180 - \angle D}{2} = \frac{\angle B}{2},\] so $BY$ bisects $\angle B$. Analogously, $CY$ also bisects $\angle C$, so the bisectors of $\angle B$ and $\angle C$ meet at $Y$ on $AD$. $\blacksquare$

Let us construct a point $T$ lying on $\overline{AB}$ such that $AD=AT$ and let $O$ be the center of the circle mentioned in the problem. Let $\angle{DTA}=\alpha.$ Then, we have $\angle{DTB}=180^\circ-\alpha$ and also \begin{align*} \angle{A} = 180^\circ-2\alpha &\implies \angle{DCB} = 2\alpha \\ &\implies \angle{DCO} = \alpha. \end{align*}Hence, $$\angle{DTO}+\angle{DCO}=180^\circ,$$so $DCOT$ is cyclic. In a similar way we can show that if we construct a point $V$ on $\overline{AB}$ such that $BC=BV,$ then $CDVO$ is cyclic. However, the circumcircle of $(COD)$ intersects $\overline{AB}$ in exactly one other point, so $T=V$ which implies that \begin{align*} AD+BC &= AT+BT \\ &= AB, \end{align*}as desired. $\square$

grobber wrote: , which means that $AD=AT$. I did not get how did we suddenly conclude this from the previous work?

Let $T$ be the point on $\overline{AB}$ such that $AD = AT$. We make the subsequent claim: Claim: $D, C, T, O$ are concyclic. Proof: The following angle chase yields: \[\angle DTA = \frac{1}{2}(180^{\circ} - \angle A) = \frac{1}{2} \angle C = \angle OCD, \]as $OC$ bisects $\angle C$ due to the tangency conditions enforced on the circle with circumcenter on $AB$. $\square$ It then follows that \[\angle BTC = \angle ODC = \frac{1}{2} \angle D, \]so that $\triangle BTC$ is isosceles. In turn, we find that $AD + BC = AT + TB = AB$, as desired. $\blacksquare$

v_Enhance 6826 posts #26VPMNov 24, 2022, 3:03 AM• 4 Y Let $T$ be the point such that $DA = AT$. Claim: $T$ lies on $(DOC)$. Proof. Because\[ \angle DCO = \frac{1}{2} \angle DCB = \frac{1}{2}(180^{\circ} - \angle BAD) = 90^{\circ} - \frac{1}{2} \angle TAD = \angle DTA. \]$\blacksquare$ [asy][asy] pair X = dir(105); pair Z = dir(160); pair Y = dir(35); pair D = 2*X*Z/(X+Z); pair A = 2/(Z+1/Z); pair B = 2/(Y+1/Y); pair C = 2*X*Y/(X+Y); filldraw(A--B--C--D--cycle, invisible, blue); draw(circumcircle(A, B, C), grey); pair O = origin; draw(arc(O, 1, 0, 180), blue); pair T = -O+2*foot(circumcenter(D, O, C), A, B); filldraw(circumcircle(D, O, C), invisible, deepgreen); draw(D--T--C, deepgreen); draw(D--O--C, lightcyan); dot("$D$", D, dir(D)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$O$", O, dir(270)); dot("$T$", T, dir(270)); /* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ X := dir 105 Z := dir 160 Y := dir 35 D = 2*X*Z/(X+Z) A = 2/(Z+1/Z) B = 2/(Y+1/Y) C = 2*X*Y/(X+Y) A--B--C--D--cycle / 0.1 lightcyan / blue circumcircle A B C / grey O 270 = origin arc O 1 0 180 / blue T 270 = -O+2*foot (circumcenter D O C) A B circumcircle D O C / 0.1 yellow / deepgreen D--T--C / deepgreen D--O--C / lightcyan */ [/asy][/asy] Reversing the previous proof on the other side gives $BC = BT$. So $AB = AT + TB = AD + BC$.

I have discussed this problem in my EGMO YouTube tutorial on my channel "little fermat" on this Video

Oops. Let $R$ be the radius of the circle and just construct the altitudes from the center to the other sides to remove the circle condition. Then what we need to prove reduces to $$\frac{R}{\tan{A}} + \frac{R}{\tan{(90 - \frac{B}{2})}} + \frac{R}{\tan{B}} + \frac{R}{\tan{(90 - \frac{A}{2})}} = \frac{R}{\sin{A}} + \frac{R}{\sin{B}}$$$$\implies \frac{1}{\tan{A}} + \tan{\frac{B}{2}} + \frac{1}{\tan{B}} + \tan{\frac{A}{2}} = \frac{1}{\sin{A}} + \frac{1}{\sin{B}} $$which after writing $\tan{\frac{A}{2}} = \frac{1 - \cos{A}}{\sin{A}}$ is obvious.