Tự vẽ hình

a,Có \(\widehat{cAe}+\widehat{cAd}=180^{o^{ }}\)(Vì kề bù)

Vì Ad là p/g \(\widehat{cAe}\Rightarrow A_1=A_2=\frac{\widehat{cAe}}{2}=\frac{120^o}{2}=60^o\)

b,Ta có:\(A_1+bAd=180^o\)(vì kề bù)

\(\Rightarrow\widehat{bAd}=120^o\)

\(\Rightarrow bAd>bAc\left(120^o>60^o\right)\)

Mà \(\widehat{bAd}=2.\widehat{bAc}\)

=>Ac là p/g \(\widehat{bAd}\)

c, có \(\widehat{cAe}+A_4=180^o\)(vì kề bù)

\(\Rightarrow A_4=60^o\)

Có:\(\widehat{bAg}+A_4=180^o\)

\(\Rightarrow\widehat{bAg}=120^o\)

Vì Ah là p/g\(\widehat{bAg}\Rightarrow A_5=\widehat{bAg}\div2=60^o\)

TA có:\(\widehat{A_1}+A_4+A_5=60^o+60^o+60^o=180^o\)

\(\Rightarrow\widehat{dAh}=180^o\)

=>2 tia Ad và Ah đối nhau

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