Xét \(\Delta\) IDC vuông tại I có:
\(\widehat{IDC}+\widehat{ICD}=180^o-\widehat{DIC}=180^o-90^o=90^o\)
mà \(\widehat{D}=2\widehat{IDC}\);\(\widehat{C}=2\widehat{ICD}\)
\(\Rightarrow\widehat{C}+\widehat{D}=2\left(\widehat{IDC}+\widehat{ICD}\right)=2\cdot90^o=180^o\)
Vì Δ`IDC` là tam giác vuông
`=> `\(\widehat{D}+\widehat{C}=90^0\)(2 góc nhon phụ nhau)
Có :`DI` là phân giác \(\widehat{ADC}\)(gt)
`=> `\(\widehat{ADI}=\widehat{IDC}=\dfrac{1}{2}\widehat{ADC}\)
`=>`\(\widehat{D}=2\widehat{IDC}\) `(1)
Lại có : `CI` Là phân giác\(\widehat{C}\)(gt)`
`=>`\(\widehat{BCI}=\widehat{ICD}=\dfrac{1}{2}\widehat{C}\)
`=>` \(\widehat{C}=2\widehat{ICD}\) `(2)`
Cộng `(1)` và `(2)`
\(\widehat{D}+\widehat{C}=2\left(\widehat{ICD}+\widehat{IDC}\right)\)
`= 2 . 90^0`
`=180^0`
