Chứng minh góc EAB = góc DAC biết AB = AC, D và E thuộc cạnh BC sao cho BD = DE = EC

\(a)Xét\Delta ABEvà\Delta ACD,tacó:\)

\(\left\{{}\begin{matrix}AE=AD\left(gt\right)\\gócAchung\\BE=DC\left(cmt\right)\end{matrix}\right.\)

\(\Rightarrow\Delta AEB=\Delta ADC\left(c-g-c\right)\)

\(\Rightarrow EAB=DAC\left(2góctươngứng\right)\)

\(\uparrow\)

\(\left\{{}\begin{matrix}BE=DE+BD\\DC=DE+EC\end{matrix}\right.\)

\(MàBD=EC\left(gt\right)\)

\(\Rightarrow BE=DC\)

\(b)A_1=A_2\left(địnhnghĩa\Delta\right)\)

\(\Rightarrow AMlàtiaphângiáccủagócDAE\)

\(\uparrow\)

\(Xét\Delta ADMvà\Delta AEM,tacó:\)

\(\left\{{}\begin{matrix}AD=AE\left(gt\right)\\AMchung\\DM=ME\left(cmt\right)\end{matrix}\right.\)

\(\Rightarrow AMD=\Delta AME\left(c-c-c\right)\)

\(\uparrow\)

\(\left\{{}\begin{matrix}DM=BM-BD\\ME=MC-EC\end{matrix}\right.\)

\(MàBM=MC\left(gt\right)\)

\(BD=EC\left(gt\right)\)

\(\Rightarrow DM=ME\)

\(c)TổnghaigócD_1vàE_1:\)

\(\Rightarrow DAE+D_1+E_1=180^o\left(tổngbagóccủatamgiác\right)\)

\(\Rightarrow60^o+D_1+E_1=180^o\)

\(\Rightarrow D_1+E_1=180^o-60^o\)

\(\Rightarrow D_1+E_1=120^o\)

\(MàD_1=E_1\left(\Delta AMD=\Delta AME\right)\)

\(\Rightarrow D_1=E_1=\dfrac{120^o}{2}=60^o\)