Giải thích các bước giải:
Bài 7:
a. Ta có : $\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^o$
$\rightarrow 80^o+40^o+\widehat{BAC}=180^o\rightarrow\widehat{BAC}=60^o$
Vì AD là phân giác $\widehat{BAC}\rightarrow\widehat{DAC}=\widehat{DAB}=\dfrac{1}{2}\widehat{BAC}=30^o$
$\rightarrow\widehat{ADC}=180^o-\widehat{ACB}-\widehat{DAC}=110^o$
b.Ta có :
$\begin{cases}AB=AE\\ \widehat{BAD}=\widehat{DAE}\\ chung \quad AD\end{cases}\rightarrow\Delta ABD=\Delta AED(c.g.c)$
c.Vì $BI$ là phân giác $\widehat{ABC}\rightarrow\widehat{ABI}=\widehat{IBC}=\dfrac{1}{2}\widehat{ABC}=40^o$
$\rightarrow\widehat{BIC}=180^o-\widehat{IBC}-\widehat{ICB}=100^o$
Theo câu b$\rightarrow\widehat{AED}=\widehat{ABC}=80^o$
$\rightarrow\widehat{BIE}+\widehat{AED}=100^o+80^o=180^o\rightarrow DE//BI$
