Giải thích các bước giải:
a.Ta có: $$\widehat{bOd}=\widehat{aOc}=75^o$$(đối đỉnh)
$$\widehat{bOc}=\widehat{aOd}=180^o-\widehat{aOc}=105^o$$
b.Ta có: $$\widehat{aOc}=\widehat{bOd}$$(đối đỉnh)
$$\widehat{aOc}+\widehat{bOd}=140^o$$
$$\to\widehat{aOc}=\widehat{bOd}=70^o$$
c.Ta có:
$$\widehat{aOc}+\widehat{cOb}+\widehat{bOd}+\widehat{dOa}=360^o$$
$$\to 2(\widehat{aOc}+\widehat{bOd})=360^o$$ vì $$\widehat{aOc}+\widehat{bOd}=\widehat{bOc}+\widehat{aOd}$$
$$\to \widehat{aOc}+\widehat{bOd}=180^o$$
Mà $$\widehat{aOc}=\widehat{bOd}$$(đối đỉnh)
$$\to \widehat{aOc}=\widehat{bOd}=90^o$$
$$\to ab\perp cd$$
$$\to \widehat{aOc}=\widehat{bOc}=\widehat{bOd}=\widehat{aOd}=90^o$$
d.Ta có:
$$\widehat{bOc}-\widehat{aOc}=10^o$$
$$\to \widehat{bOc}=\widehat{aOc}+10^o$$
Mà $$\widehat{aOc}+\widehat{bOc}=180^o$$
$$\to \widehat{aOc}+(\widehat{aOc}+10^o)=180^o$$
$$\to 2\widehat{aOc}=170^o$$
$$\to \widehat{aOc}=85^o$$
$$\to \widehat{bOd}=\widehat{aOc}=85^o,\widehat{bOc}=\widehat{aOd}=180^o-\widehat{aOc}=95^o$$
e.Ta có:
$$\widehat{aOc}+\widehat{bOc}=180^o$$
$$\to \widehat{aOc}+2\widehat{aOc}=180^o$$
$$\to 3\widehat{aOc}=180^o$$
$$\to \widehat{aOc}=60^o$$
$$\to \widehat{bOc}=120^o$$
$$\to \widehat{bOd}=\widehat{aOc}=60^o,\widehat{aOd}=\widehat{bOc}=120^o$$
