Đáp án+Giải thích các bước giải:
`Gọi IF∩CD={N}`
`text{-Theo định lí về góc noài của tam giác:}`
+)`text{ΔINE có:}``\hat{FIE}=\hat{INE}+\hat{IEN}`
hay `\hat{FIE}=\hat{FNE}+\hat{AED}/2`
+)`text{ΔDNF có:}``\hat{FNE}=\hat{D}+\hat{DFN}`
hay`\hat{FNE}=\hat{D}+\hat{DFC}/2`
⇒`\hat{FIE}=\hat{D}+\hat{DFC}/2+\hat{AED}/2`
hay `\hat{FIE}=\hat{D}+(\hat{DFC}+\hat{AED})/2``(1)`
`text{-ΔADE có:}`
`\hat{AED}=180^0-(\hat{D}+\hat{DAE})`
`text{-ΔDFC có:}`
`\hat{DFC}=180^0-(\hat{D}-\hat{DCF})`
⇒`\hat{AED}+\hat{DFC}=360^0-(2\hat{D}+\hat{DAE}+\hat{DCF})`
mà tứ giác `ABCD` có:`\hat{DAB}+\hat{ABC}+\hat{BCD}+\hat{D}=360^0`
hay`\hat{DAE}+\hat{ABC}+\hat{DCF}+\hat{D}=360^0`
⇒`\hat{AED}+\hat{DFC}=\hat{DAE}+\hat{ABC}+\hat{DCF}+\hat{D}-(2\hat{D}+\hat{DAE}+\hat{DCF})`
`=\hat{ABC}-\hat{D}`
`text{Thay vào (1) ta có:}`
`\hat{FIE}=\hat{D}+(\hat{ABC}-\hat{D})/2`
`=\hat{D}/2+(\hat{ABC}-\hat{D})/2`
`=(\hat{ABC}+\hat{D})/2`
hay `\hat{FIE}=(\hat{ABC}+\hat{ADC})/2`(đpcm)
`____________________`
Chúc bạn học tốt!!!
`#Rùa~ ~ ~`
