Exercițiul 976

E.976. Cretu

Alin Crețu, MateMaraton, 31.05.2026
Soluție
Soluție:

Construim:

N∈DH   a.ıˆ.   HN=HAM∈DH   a.ıˆ.   DM=DAL∈DG   a.ıˆ.   DL=DNK∈CG   a.ıˆ.   CK=CA}⇒ triunghiurile HNA,DAM,DLN,BKM,CKA sunt isoscele. \begin{rcases} N \in DH \text{ ~~a.î.~~ } HN=HA \\ M \in DH \text{ ~~a.î.~~ } DM=DA \\ L \in DG \text{ ~~a.î.~~ } DL=DN \\ K \in CG \text{ ~~a.î.~~ } CK=CA \end{rcases} \Rightarrow \text{ triunghiurile } HNA,DAM,DLN, BKM, CKA \text{ sunt isoscele.}

2x^=CHB^=180−HCB^−B^=KCA^−B^=180−2z^−B^2y^=2BKM^−2z^=180−B^−2z^}⇒x^=y^⇒NMAK inscriptibil. \begin{rcases} 2\widehat{x} = \widehat{CHB} = 180-\widehat{HCB}-\widehat{B}=\widehat{KCA}-\widehat{B}=180-2\widehat{z}-\widehat{B} \\ 2\widehat{y}=2\widehat{BKM}-2\widehat{z}=180-\widehat{B}-2\widehat{z} \end{rcases} \Rightarrow \boxed{\widehat{x}=\widehat{y}} \Rightarrow NMAK \text{ inscriptibil}.

NMALNMAL este trapez isoscel, deci și NMALNMAL este inscriptibil ⇒N,M,A,K,L\Rightarrow N,M,A,K,L conciclice ⇒KLA^=KMA^=u^.\Rightarrow \boxed{\widehat{KLA}=\widehat{KMA}=\widehat{u}}.

Dar u^=KMB^−AMD^=180−B^2−180−d^2=d^−B^2=v^2⇒△LGK\widehat{u}=\widehat{KMB}-\widehat{AMD}=\dfrac{180-\widehat{B}}{2} - \dfrac{180-\widehat{d}}{2} = \dfrac{\widehat{d}-\widehat{B}}{2} = \dfrac{\widehat{v}}{2} \Rightarrow \triangle LGK isoscel ⇒GL=GK.\Rightarrow \boxed{GL=GK}.

În concluzie, (DH+HA)−(CG+GA)=(DH+HN)−(CK+KG+GA)=DL−(AE+GL+GA)=ED=x.(DH+HA) - (CG+GA) = (DH+HN) - (CK+KG+GA) =DL-(AE+GL+GA) = ED=x.