Exercițiul 807

E.807. AB<AC, BM=MC, AEMDAB<AC,~ BM=MC,~ AEMD paralelogram, AA și FF puncte de tangență.
Demonstrați că ADF^=AEN^.\widehat{ADF} = \widehat{AEN}.

Alin Crețu, MateMaraton, 04.11.2025
Soluție
Soluție:

Fie OO - centrul cercului (A,B,C).(A,B,C).

  • BM=MC⇒OMD^=90°.BM=MC \Rightarrow \boxed{\widehat{OMD}=90\degree}.
  • DA,DFDA,DF tangente ⇒DA=DF, DAO^=DFO^=90°, D1^+D2^=D3^.\Rightarrow \boxed{DA=DF},~ \boxed{\widehat{DAO}=\widehat{DFO}=90\degree}, ~ \boxed{\widehat{D_1}+\widehat{D_2} = \widehat{D_3}}.
  • EM∥AD⇒ALM^=90°.EM \parallel AD \Rightarrow \boxed{\widehat{ALM}=90\degree}.

A^=F^=M^=90°⇒A,D,F,M,O conciclice\widehat{A}=\widehat{F}=\widehat{M} = 90\degree \Rightarrow \boxed{A,D,F,M,O \text{ conciclice}}.

ADFM inscriptibil ⇒M1^+M2^=ADF^=2(D1^+D2^)DFMO inscriptibil ⇒M1^=D1^+D2^}⇒M1^=M2^⇒MF=MN. \begin{rcases} ADFM \text{ inscriptibil } \Rightarrow \widehat{M_1} + \widehat{M_2} = \widehat{ADF} = 2(\widehat{D_1} + \widehat{D_2}) \\ DFMO \text{ inscriptibil } \Rightarrow \widehat{M_1} = \widehat{D_1} + \widehat{D_2} \end{rcases} \Rightarrow \widehat{M_1} = \widehat{M_2} \Rightarrow \boxed{MF=MN}.

DFM^=90°+F2^EMN^=90°+A2^}⇒F2^=A2^=D2^DFM^=EMN^. \begin{rcases} \widehat{DFM} = 90\degree + \widehat{F_2} \\ \widehat{EMN} = 90\degree + \widehat{A_2} \end{rcases} \overset{\widehat{F_2} = \widehat{A_2} = \widehat{D_2}}{\Rightarrow} \boxed{\widehat{DFM} = \widehat{EMN}}.
△DFM≡△EMN (L.U.L.) ⇒D^1=E^1EMDA paralelogram ⇒D2^+D3^=AEM^}⇒ADF^=AEM^. \begin{rcases} \triangle DFM \equiv \triangle EMN \text{ (L.U.L.) } \Rightarrow \widehat D_1 = \widehat E_1 \\ EMDA \text{ paralelogram } \Rightarrow \widehat{D_2} + \widehat{D_3} = \widehat{AEM} \end{rcases} \Rightarrow \boxed{\widehat{ADF} = \widehat{AEM}}.