Question

Answer 1

a) Ta có: \(\angle BHE=\angle BFE=90\Rightarrow BEHF\) nội tiếp

b) Ta có: \(\angle EFB=\angle EMB=90\Rightarrow EFMB\) nội tiếp

\(\Rightarrow B,E,H,F,M\) cùng thuộc 1 đường tròn

Ta có: \(\angle OEB=\dfrac{180-\angle BOE}{2}=90-\dfrac{1}{2}\angle BOE=90-\angle BAE=\angle AEF\)

\(\Rightarrow\angle AEO=\angle BEF\)

mà \(\left\{{}\begin{matrix}\angle AEO=\angle HBC\left(HEBMnt\right)\\\angle BEF=\angle CMF\left(BEFMnt\right)\end{matrix}\right.\)

\(\Rightarrow\angle HBC=\angle CMF\) \(\Rightarrow FM\parallel BH\) mà \(BH\bot AE\Rightarrow FM\bot AE\)

c) Kẻ \(EQ'\bot AC\)

Ta có: \(\angle EFA+\angle EQ'A=90+90=180\Rightarrow AFEQ'\) nội tiếp 

\(\Rightarrow\angle AFQ'=\angle AEQ'=90-\angle Q'AE=90-\angle EBC=\angle MEB=\angle MFB\)

mà A,F,B thẳng hàng \(\Rightarrow Q',F,M\) thẳng hàng \(\Rightarrow Q\equiv Q'\)