Question

Answer 1

a: HI là phân giác của góc AHB

=>\(\hat{AHI}=\frac12\cdot\hat{AHB}=\frac12\cdot90^0=45^0\)

HJ là phân giác của góc AHC

=>\(\hat{AHJ}=\frac12\cdot\hat{AHC}=\frac{90^0}{2}=45^0\)

\(\hat{IHJ}=\hat{IHA}+\hat{JHA}\)

\(=45^0+45^0=90^0\)

b: Xét ΔAHB có \(\hat{AHB}+\hat{ABH}+\hat{HAB}=180^0\)

=>\(\hat{AHB}+\hat{ABH}=180^0-\hat{HAB}\)

=>\(2\left(\hat{IBH}+\hat{IHB}\right)=180^0-\hat{HAB}\)

=>\(\hat{IBH}+\hat{IHB}=90^0-\frac12\cdot\hat{HAB}\)

Xét ΔIBH có \(\hat{IBH}+\hat{IHB}+\hat{BIH}=180^0\)

=>\(\hat{BIH}=180^0-90^0+\frac12\cdot\hat{HAB}=90^0+\frac12\cdot\hat{HAB}\)

XétΔAHC có \(\hat{AHC}+\hat{ACH}+\hat{HAC}=180^0\)

=>\(\hat{AHC}+\hat{ACH}=180^0-\hat{HAC}\)

=>\(2\left(\hat{JHC}+\hat{JCH}\right)=180^0-\hat{HAC}\)

=>\(\hat{JHC}+\hat{JCH}=90^0-\frac12\cdot\hat{HAC}\)

Xét ΔJCH có \(\hat{JCH}+\hat{JHC}+\hat{HJC}=180^0\)

=>\(\hat{HJC}=180^0-90^0+\frac12\cdot\hat{HAC}=90^0+\frac12\cdot\hat{HAC}\)

\(\hat{BIH}+\hat{CJH}\)

\(=90^0+\frac12\cdot\hat{HAC}+90^0+\frac12\cdot\hat{HAB}=180^0+\frac12\cdot90^0=225^0\)