\(\widehat{ABC}+\widehat{ACB}=180^0-\widehat{BAC}\)
=>\(\widehat{IBC}+\widehat{ICB}=90^0-\dfrac{1}{2}\widehat{BAC}\)
\(\widehat{BIC}=180^0-\left(90^0-\dfrac{1}{2}\cdot\widehat{BAC}\right)=\dfrac{1}{2}\widehat{BAC}+90^0\)
Xét ΔADI vuông tại D và ΔAEI vuông tại E có
AI chung
\(\widehat{DAI}=\widehat{EAI}\)
Do đo: ΔADI=ΔAEI
Suy ra: AD=AE
hay ΔADE cân tại A
=>\(\widehat{ADE}=\dfrac{180^0-\widehat{A}}{2}\)
\(\widehat{BIC}+\widehat{ADE}=\dfrac{1}{2}\cdot\widehat{BAC}+90^0-\dfrac{1}{2}\cdot\widehat{BAC}+90^0=180^0\)