Problema #5 – Sumă de unghiuri în triunghi

Fie \;ABC \;un \;triunghi \;echilateral\; și\; punctul \;D\; astfel \;încât\;\;\;\;\;\; \\BD = BC, \;m(\measuredangle CBD) = 20^{\circ}\; iar\; punctele \;A\; şi \;D\; sunt \;situate\;\;\\de \;o \;parte \;și \;de\; alta \;a\; dreptei \;BC.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \\Fie M \in 	[ \,BD	] \, astfel\; încât\; BM = DC.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \\Aflați \;m(\measuredangle DMC).\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;

\left.
    \begin{array}{ll}
m(\measuredangle DBC)\;=\; 20^{\circ} (din\;ipoteză)\\\;\\
m(\measuredangle ABC)\;=\; 60^{\circ}(△ ABC\; △ echilateral) \\
\;\;\;\;\;\;\;\;\\m(\measuredangle ABM)\; = \; m(\measuredangle ABC) + m(\measuredangle CBM)\\\;\\
  \end{array}
\right \}  
\implies\;\\\; \\ \implies m(\measuredangle ABM)\; = 60^{\circ} \;+\;20^{\circ} \implies\;m(\measuredangle ABM)\; = 80^{\circ}\\                     \;\\ \;\\

BC = BD\;(din\;ipoteza) {\implies}△ CBD\; △isoscel\implies\;\;\;\;\;\;\;\;\;\;\;\;\;\\\;\\
\left.
    \begin{array}{ll}
\implies \;m(\measuredangle BCD)\;=\;m(\measuredangle CDB)\; \\ \\
\;\;\;\;\;\;\;\;\;\;m(\measuredangle DBC)\;=\; 20^{\circ}\;(din\; ipoteză)
\\ \\\;\;\;\;\;\;\;\;\;\;m(\measuredangle CBD)\; + \; m(\measuredangle CDB) + m(\measuredangle DCB) = 180^{\circ}\\\;\\ 
  \end{array}
\right \}  
\implies\;   \\\;\\\implies2 \cdot m(\measuredangle CDB) + 20^{\circ}  = 180^{\circ} \implies\; 2 \cdot m(\measuredangle CDB)  = 160^{\circ}  \implies\;\\\;\\\implies\; m(\measuredangle CDB)  = 80^{\circ} 

\left.
    \begin{array}{ll}

\left.
    \begin{array}{ll}
m(\measuredangle ABM)\; = 80^{\circ} \\ \;\\ m(\measuredangle CDB)  = 80^{\circ} 
  \end{array}
\right \}  \implies m(\measuredangle ABM)\; =\;m(\measuredangle CDB)  
\\\;\\\\\;\\
\left.
    \begin{array}{ll}
△ ABC\; △ echilateral\implies AB = BC\\\;\\
BC = BD \;(din\; ipoteză)
  \end{array}
\right \}  \implies AB = BD\;\\\;\\\\\;\\
BM = CD \;(din\;ipoteză)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;
  \end{array}
\right \}  \overset{LUL} {\implies} \\ \; \\\ \;\\\overset{LUL} {\implies}\;△ABM\;{\displaystyle \equiv }\;△BDC\implies \\ \;\\ \;\\
\left.
    \begin{array}{ll}
 \implies m(\measuredangle AMB)\; = m(\measuredangle BCD) \\\;\\
\;\;\;\;\;\;\;\;\;m(\measuredangle BCD) = 80^{\circ}
  \end{array}

\right \}  \implies m(\measuredangle AMB) = 80^{\circ}  \\\;\\\;\\
\left.
    \begin{array}{ll}
m(\measuredangle AMB) = 80^{\circ}\\\;\\
m(\measuredangle ABM) = 80^{\circ}
  \end{array}
\right \}  \implies  △BAM\;△isoscel

\left.
    \begin{array}{ll}

△ABM\;{\displaystyle \equiv }\;△BDC\implies m(\measuredangle BAM) = m(\measuredangle DBC) \\\;\\
m(\measuredangle CBD) = 20^{\circ} (ipoteză)
  \end{array}
\right \} \implies \\\;\\\;\\ 

\left.
    \begin{array}{ll}
\implies m(\measuredangle BAM) = 20^{\circ}
\\\;\\
\;\;\;\;\;\;\;\;\;m(\measuredangle BAC)\;=\; 60^{\circ}(△ ABC\; △ echilateral) 
 \\\;\\
\;\;\;\;\;\;\;\;\;m(\measuredangle BAC) = m(\measuredangle BAM) \;+\; m(\measuredangle MAC)
  \end{array}
\right \}  \implies \\\;\\\implies m(\measuredangle MAC) = m(\measuredangle BAC)\;-\;m(\measuredangle BAM) \implies \\\;\\\implies m(\measuredangle MAC)\; =\; 60^{\circ}\;-\;20^{\circ}\implies m(\measuredangle MAC)\; =\; 40^{\circ}

\left.
    \begin{array}{ll}

△BAM\;△isoscel\implies AB \;= \;AM \\\;\\
△ABC\;△echilateral\implies AB\; = \;AC
  \end{array}
\right \} \implies \\\;\\\;\\ \implies AM=AC \implies △MAC\;△isoscel \implies \\\;\\
\left.
    \begin{array}{ll}
\implies m(\measuredangle AMC) = m(\measuredangle ACM)\;
\\\;\\
\;\;\;\;\;\;\;\;\;m(\measuredangle MAC)\; =\; 40^{\circ}\\\;\\

\;\;\;\;\;\;\;\;\;m(\measuredangle MAC) + m(\measuredangle ACM) \;+\; m(\measuredangle CMA)  \; =\; 180^{\circ}\\\;\\
  \end{array}
\right \}  \implies \\\;\\\implies 40^{\circ} + 2 \cdot m(\measuredangle AMC) = 180^{\circ} \implies 2 \cdot m(\measuredangle AMC) = 140^{\circ}\implies \;\\\\\;\\\implies  m(\measuredangle AMC)\; =\; 70^{\circ}

\left.
    \begin{array}{ll}

m(\measuredangle BMA) + m(\measuredangle AMC) \;+\; m(\measuredangle CMD)  \; =\; 180^{\circ}\\\;\\
 m(\measuredangle AMB) \;= \;80^{\circ}\;\\\;\\
m(\measuredangle MAC)\; =\; 70^{\circ}
  \end{array}
\right \}  \implies \\\;\;\\\implies m(\measuredangle CMD)   =180^{\circ}-m(\measuredangle BMA)-m(\measuredangle AMC) \;\implies\\\;\\\implies m(\measuredangle CMD)   =\;180^{\circ}-\;80^{\circ}\;\;-70^{\circ}\;\;\implies\\\;\\\implies m(\measuredangle CMD)   =\;30^{\circ}