xét tứ giác ABCD có:
\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)
mà \(\widehat{A}=\widehat{B}=\widehat{C}=90^0\)
\(\Rightarrow90^0+90^0+90^0+\widehat{D}=360^0\)
\(\Rightarrow270^0+\widehat{D}=360^0\)
\(\Rightarrow\widehat{D}=90^0\)
\(\Rightarrow AD\perp DC\left(đpcm\right)\)
DA ⊥ AB, CB ⊥ AB => DA // CB
Mà DC ⊥ BC => CD ⊥ AD
Ta có:
DA ⊥ AB, CB ⊥ AB => DA // CB
Mà DC ⊥ BC => CD ⊥ AD
=>đpcm