`#huy`
`a)` Theo đề ta có:
`\hat(A);\hat(B);\hat(C);\hat(D)=1;2;3;4`
`=(\hat(A))/1=(\hat(B))/2=(\hat(C))/3=(\hat(D))/4=(\hat(A)+\hat(B)+\hat(C)+\hat(D))/(1+2+3+4)`
`=(360^0)/10=36^0` (vì `\hat(A)+\hat(B)+\hat(C)+\hat(D)=360^0`)
`=>\hat(A)=36^0 . 1=36^0`
`=>\hat(B)=36^0 . 3=72^0`
`=>\hat(C)=36^0 . 3=108^0`
`=>\hat(D)=36^0 . 4=144^0`
`b)` Do `{:(\hat(A)+\hat(D)=180^0),(\hat(B)+\hat(C)=180^o):}}=>AB////CD`
`c)\hat(EDC)=180^0-\hat(D_1)-\hat(C_1)`
`=180^0-\hat(A_1)-\hat(B_1)` (do `AB////CD=>\hat(A_1)=\hat(D_1);\hat(B_1)=\hat(C)`)
`=180^0-36^0-72^0`
`=72^0`