N.T.TUAN wrote: Given an isosceles triangle $ ABC$ with $ AB = BC$. A point $ M$ is chosen inside $ AB$ such that $ \angle AMC = 2\angle ABC$ . A point $ K$ lies on segment $ AM$ such that $ \angle BKM =\angle ABC$. Prove that $ BK = KM+MC$. Maybe you mean $ M$ is chosen inside triangle $ ABC$.

Nice,but easy: Denote $ CM\cap BK = N,\angle ABC =\alpha$. 1.Since $ \angle BKM =\angle NKM =\alpha$ and $ \angle AMC = 2\alpha$ then $ \angle KNM =\alpha\Rightarrow NM = MK$. 2.$ \angle NCB =\angle KNC-\angle NBC =\alpha-\angle NBC =\angle KBA$. 3.$ \angle BAM =\alpha-\angle KBA$.Now using 2.:$ \angle BAM =\alpha-\angle NCB = NBC$. 4.Using 2.,3.and $ AB = BC$ we get that $ \triangle BAK =\triangle CNB\Rightarrow BK = NC = CM+MN = CM+MK$

N.T.TUAN wrote: Given an isosceles triangle $ ABC$ with $ AB = BC$. A point $ M$ is chosen inside $ ABC$ such that $ \angle AMC = 2\angle ABC$ . A point $ K$ lies on segment $ AM$ such that $ \angle BKM =\angle ABC$. Prove that $ BK = KM+MC$. Proof (similarly with the nice Erken's proof). Denote $ m(\widehat{ABC})=\beta$, $ N\in CM\cap BK$, the circumcenter $ O$ and the orthocenter $ H$ for $ \triangle ABC$. $ 1\blacktriangleright\{\begin{array}{c}m(\widehat{AMC})=2\cdot m(\widehat{ABC})=2\beta\\\ m(\widehat{BKM})=m(\widehat{ABC})=\beta\end{array}\|$ $ \implies$ $ m(\widehat{KNM})=m(\widehat{AMC})-m(\widehat{BKM})=2\beta-\beta=\beta$ $ \implies$ $ \widehat{NKM}\equiv\widehat{KNM}$ $ \implies$ the triangle $ KMN$ is $ M$- isosceles $ \implies$ $ \boxed{MK=MN}$. $ 2\blacktriangleright\{\begin{array}{c}m(\widehat{ABK})+m(\widehat{CBK})=\beta=m(\widehat{CNK})=m(\widehat{BCN})+m(\widehat{CBK})\implies\widehat{ABK}\equiv\widehat{BCN}\\\ m(\widehat{ABK})+m(\widehat{BAK})=m(\widehat{NKM})=\beta =m(\widehat{ABK})+m(\widehat{CBN})\implies\widehat{BAK}\equiv\widehat{CBN}\end{array}\|$ $ \implies$ $ \{\begin{array}{c}AB=BC\\\ \widehat{ABK}\equiv\widehat{BCN}\\\ \widehat{BAK}\equiv\widehat{CBN}\end{array}\|$ $ \implies$ $ \boxed{\triangle\ ABK\equiv\triangle\ BCN}$ $ \implies$ $ \{\begin{array}{c}AK=BN\\\ BK=CN\\\ \widehat{AKB}\equiv\widehat{BNC}\end{array}$. Therefore, $ BK=CN=NM+MC=KM+MC\implies\boxed{BK=KM+MC}$. $ 3\blacktriangleright m(\widehat{AOC})=m(\widehat{AMC})=2\beta\implies$ the quadrilateral $ AOMC$ is cyclically $ \implies$ $ m(\widehat{OMN})=m(\widehat{OAC})=90^{\circ}-\beta$ $ \implies$ $ OM\perp BK$. $ 4\blacktriangleright\ldots\ldots\implies$ the quadrilaterals $ ABOK$ and $ BNOC$ are cyclically. $ 5\blacktriangleright\ldots\ldots$ $ \implies$ the point $ O$ is the incenter of the isosceles triangle $ MNK$. I"ll come back here later !