can any one post the official solution ?

Solution. easy! Choose a point $K$ on $\overline{BC}$ such that $BD=BK$. Note that $ADKB$ is cyclic. Hence $AD=DK$, moreover, by some angle chasing we have $DK=KC$. Thus $AD=KC$. And we are done. $\blacksquare$

ayan.nmath wrote: Solution. Choose a point $K$ on $\overline{BC}$ such that $BD=BK$. Note that $ADKB$ is cyclic. Hence $AD=DK$, moreover, by some angle chasing we have $DK=KC$. Thus $AD=KC$. And we are done. $\blacksquare$ nice solution

Oops, when I tried to make a synthetic solution I extended it, no wonder it didn't work. Will using Sine Law give 10 points? It would even fit in 1 page.

Note that $\angle A=100^{\circ}$. Let $D'$ be a point on $BC$ such that $BD=BD'$. Hence, $\angle BDD'=\angle BD'D=80^{\circ}$. Also, $\angle DD'C=100^{\circ}$ and $\angle D'DC=40^{\circ}$ so $DD'=D'C$. It suffices to prove that $AD=DD'$. Note that $\angle BAD+\angle DD'B=100^{\circ}+80^{\circ}=180^{\circ}$ so $ADD'B$ is cyclic so $$\angle D'AD=\angle DBD'=20^{\circ}=\angle ABD=\angle ADD'$$so $AD=DD'$, as desired. $\blacksquare$

see figure:

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I am very bad at making synthetic observations, so here's my try at Trig-bashing Let $BD$ be produced to $A'$, such, $AD=A'D$ and Let $AB=b=AC$ and $BC=a$ Therefore, $BD=\frac{2ab}{a+b} \cos 20^{\circ}$ and using angle bisector theorem, easy to find out, $A'D=\frac{b^2}{a+b}$ Let's assume $BD+AD=BC \implies \frac{2ab}{a+b} \cos 20^{\circ}+\frac{b^2}{a+b} =a \implies 2ab\cos 20^{\circ} +b^2=a^2+ab$ Now sine rule gives us, $\boxed{a=2b\cos 40^{\circ}} \implies 4b^2\cos 20^{\circ} \cos 40^{\circ}+b^2 =4b^2 \cos ^2 40^{\circ} +2b^2 \cos 40^{\circ} $ Hence, $$4\cos 20^{\circ} \cos 40^{\circ}+\cos 60^{\circ}=2\cos ^2 40^{\circ}+\cos 40^{\circ} \implies 2\cos 40^{\circ} \cdot 2\sin 10^{\circ} \sin 30^{\circ}=2\sin 10^{\circ} \sin 50^{\circ} \implies \sin 30^{\circ}=\frac{1}{2} \text{, which is obviously true!!}$$Hence, $\boxed{BD+AD=BC}$

my solution posted here before, without words (no cyclic needed)

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Use sine rule extensively. We write $AD=\frac{AB\sin20}{\sin60}, BD=\frac{AB\sin100}{\sin40}$ and $BC=\frac{AB\sin100}{\sin40}$. Thus, we obtain a trigonometric equation which is easy to solve if we apply the identity of $ \sin A+\sin B=2 \sin {\frac{A+B}{2}} \cos {\frac{A-B}{2}}.$ $

Does anybody knows who proposed this problem?

Also, it seemed me too easy for a sort list proplem... (Sorry for double posting but I can't edit from my tablet...)

[asy][asy] size(13cm); import olympiad; pair B = (0, 0); pair C = (5, 0); pair A = (2.5, 2); draw(A--B--C--A); label("$A$", A, N); label("$B$", B, W); label("$C$", C, E); pair E = bisectorpoint(A, B, C); pair D = extension(B, E, A, C); draw(B--D); label("$D$", D, NE); [/asy][/asy] Let's do some trig! But lets do some very easy angle chasing before we do that. We have that $\angle{ABD}=\angle{DBC}=20$ and we know that $\angle{ACB}=40$ so $\angle{BDC}=120$. Then we know that $\angle{ADB}=60, \angle{BAC}=100$. Now notice by LoS we have \begin{align*} BD&=\frac{AB \sin(100)}{\sin(60)} \\ DA&=\frac{AB \sin(20)}{\sin(60)} \\ BC&=\frac{AB \sin(100)}{\sin(40)} \end{align*} Now let's try to compute \[ BD+DA=\frac{AB(\sin(100)+\sin(20))}{\sin(60)}. \]From sum to product identities we have $\sin(100)+\sin(20)=2\sin(60)\cos(40)$. So then \[ BD+DA=\frac{AB(\sin(100)+\sin(20))}{\sin(60)}=2\cos(40)=2\sin(50) \]Notice the $2$ and the $\sin(\theta)$ which makes us think of the sine double angle identity. So we get \[ \sin(100)=\sin(50+50)=2\sin(50)\cos(50)=2\sin(50)\sin(40) \]Now we see that $BD+DA=2\sin(50)=\frac{\sin(100)}{\sin(40)}=BC$ $\blacksquare$

Consider a point F on BC such that CF=AD .by angle bisector theorem we know AB/BC=AD/DC replacing lengths we know AC/BC=CF/DC or AC/CF=BC/DC hence DFC and ABC are similar and we get our result

Select a point $E$ on $ BC$ such that$ BD=BE $. Also $ AB = AC $. $ ABED $ comes out to be cyclic, hence $ \angle DAE= \angle DBE= 20^ \circ $. similarly $ \angle DEA = \angle ABD = 20^\circ $. So $ AD = DE $. $ \angle CDE= \angle DAE + \angle DEA =40^ \circ $, which follows $ \angle DCE = \angle CDE $ hence $ DE=CE$ and thus $ AD= EC $. So $ BC = BE + EC = BD + AD $ $\blacksquare$

βλεπε σχήμα

Let $X$ on $BC$ be a point such that $BD = BX$, let $Y$ on $BC$ be a point such that $\angle XDY = 20^\circ$, and let $Z$ on $BD$ be a point such that $DA = DZ$. It suffices to show $XC = AD$. $\angle A = 100^\circ \implies D = 60^\circ \implies \triangle ADZ$ is equilateral so $AZ = AD$. $\angle DXB = \frac12 (180 - 20) = 80 \implies \angle DYX = 180^\circ - 20^\circ - 80^\circ = 80^\circ \implies DX = DY$. $\angle BDX = \angle DXB = 80^\circ \implies \angle XDC = 180^\circ - 60^\circ - 80^\circ = 40^\circ \implies DX = XC$ $\angle YDC = \angle ZAD = 60 ^\circ \implies AZ \parallel DY$. Furthermore, $\angle DYB = \angle DAB = 100^\circ \implies \triangle ABD \cong \triangle YBD$ so $BY = BA$. Thus $\triangle BAZ \cong \triangle BYZ \implies AD = AZ = ZY$. Hence $ADZY$ is a rhombus so by combining all the equalities $AD = AZ = DY = DX=XC$.

Let $E$ be the point on segment $BC$ such that $BD = BE$. Then, $$\angle BED = \frac{180^{\circ} - \frac{40^{\circ}}{2}}{2} = 80^{\circ} = 180^{\circ} - \angle BAD$$so $ABED$ is cyclic, i.e. $ABC \sim EDC$. Now, the Angle Bisector Theorem yields $$\frac{DA}{DC} = \frac{BA}{BC} = \frac{DE}{DC} = \frac{EC}{DC}$$which implies $DA = EC$. To finish, we note $$BD + DA = BE + EC = BC$$as desired. $\blacksquare$ Remark: The desired length condition intrinsically motivates the construction of $E$.