a) Có: $\\$
$\left.\begin{matrix} \text{HI là phân giác $\widehat{AHB}$ (gt) }\\\text{HJ là phân giác $\widehat{AHC}$ (gt)}\\\text{ $\widehat{AHB}$ và $\widehat{ABC}$ là 2 góc kề bù}\end{matrix}\right\}\text{⇒}$ $\widehat{ABC}$ + $\widehat{ABC}$ = $90^{o}$ $\text{(Hai tia phân giác của hai góc kề bù)}$ $\\$ $\text{=> $\widehat{IHJ}$ = $90^{o}$ }$
b) Xét ΔHIB có :
$\widehat{HIB}$ + $\widehat{IBH}$ + $\widehat{BHI}$ = $180^{o}$
⇒ $\widehat{IBH}$ = $180^{o}$ - $\widehat{BHI}$ + $\widehat{IBH}$
mà $\widehat{BHI}$ = $\dfrac{\widehat{AHB}}{2}$ và $\widehat{IBH}$ = $\dfrac{\widehat{ABH}}{2}$
=> $\widehat{IBH}$ = $180^{o}$ - $\dfrac{\widehat{AHB} + \widehat{ABH}}{2}$
mà $\widehat{AHB}$ + $\widehat{ABH}$ = $180^{o}$ - $\widehat{HAB}$
=> $\widehat{HIB}$ = $180^{o}$ - $\dfrac{180^{o} - \widehat{HAB} }{2}$
=> $\widehat{HIB}$ = $180^{o}$ - $90^{o}$ + $\dfrac{\widehat{HAB}}{2}$
=> $\widehat{HIB}$ = $90^{o}$ + $\dfrac{\widehat{HAB}}{2}$
CMTT: $\widehat{HJC}$ = $90^{o}$ + $\dfrac{\widehat{HAC}}{2}$
=> $\widehat{HJC}$ + $\widehat{HIB}$ = $90^{o}$ + $\dfrac{\widehat{HAC}}{2}$ + $90^{o}$ + $\dfrac{\widehat{HAB}}{2}$
= $180^{o}$ + $\dfrac{\widehat{ABC}}{2}$
= $180^{o}$ + $45^{o}$ = $225^{o}$
@tryphena
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