giúp mình với ạ làm ý nào bài 4 cũng dc ạ

Giải thích các bước giải:

1.Ta có:

$\widehat{BAH}=90^o-\hat B=\widehat{ACB}$

$\widehat{CAH}=90^o-\hat C=\widehat{ABC}$

2.Gọi $AH\cap CK=D$

$\to\widehat{DAK}=\dfrac12\widehat{BAH}=\dfrac12\widehat{ACH}=\widehat{DCH}$

$\to \widehat{DKA}=180^o-\widehat{DAK}-\widehat{ADK}=180^o-\widehat{DCH}-\widehat{HDC}=\widehat{DHC}=90^o$

$\to AK\perp KC$

3.Gọi $AH\cap BL=E$

Ta có: $AL\perp BL$

$\to \widehat{LAE}=90^o-\widehat{AEL}=90^o-\widehat{BEH}=\widehat{EBH}=\dfrac12\widehat{ABC}=\dfrac12\widehat{CAH}$

$\to AL$ là phân giác $\widehat{HAC}$

4.Ta có:  

$\widehat{BIC}$

$=\widehat{KIL}$

$=\widehat{AIK}+\widehat{AIL}$

$=(90^o-\widehat{IAK})+(90^o-\widehat{IAL})$

$=180^o-(\widehat{IAK}+\widehat{IAL})$

$=180^o-(\dfrac12\widehat{BAH}+\dfrac12\widehat{CAH})$

$=180^o-\dfrac12\widehat{BAC}$

$=180^o-\dfrac12\cdot 90^o$

$=135^o$