Question

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Answer 1

$\displaystyle\it\\\Delta ABE~este~isoscel~de~baza~[BE] \implies [AB]=[AE],~m(\measuredangle ABE)=m(\measuredangle AEB).\\avem,~[BC]=[ED].\\m(\measuredangle ABC)=m(\measuredangle AED),~dar: \left \{ {{m(\measuredangle ABC)=m(\measuredangle ABE)+m(\measuredangle EBC).} \atop {m(\measuredangle AED) =m(\measuredangle AEB)+m(\measuredangle DEB)}} \right. \implies\\cum~m(\measuredangle ABC)=m(\measuredangle AED)~si~m(\measuredangle ABE)=m(\measuredangle AEB) \implies$

$\displaystyle\it\\m(\measuredangle EBC)=m(\measuredangle DEB).$

$\displaystyle\it\\a)~comparam~triunghiurile~ABC~si~AED.\\avem~:~[AB]=[AE],~m(\measuredangle AED)=m(\measuredangle AED)~iar~[BC]=[DE] \stackrel{L.U.L}\Longrightarrow \\\Delta ABC \equiv \Delta AED \implies [AC]=[AD] \implies \Delta ACD ~este~isoscel.$

$\displaystyle\it\\b)~din~congruenta~triunghiurilor~de~mai~sus~\implies m(\measuredangle BAC)=m(\measuredangle EAD)~(*). \\cum~m\measuredangle(EAC)=m(\measuredangle EAD)+m(\measuredangle DAC).\\iar,~m(\measuredangle BAD)=m(\measuredangle BAC)+m(\measuredangle DAC).\\din~(*) \implies evident,~m(\measuredangle EAC)=m(\measuredangle BAD).$

$\displaystyle\it\\c)~comparam~triunghiurile~BAD~si~EAC.\\m(\measuredangle BAD)=m(\measuredangle EAC),~[AB]=[AE],~[AD]=[AC] \stackrel{L.U.L}\Longrightarrow \Delta BAD\equiv \Delta EAC,~\\de~unde~[BD]=[EC].$