a)
Xét \(ΔAHD\):
\(HA=HD\left(gt\right)\)
\(\widehat{AHD}=90^o\)
\(⇒ΔAHD\) vuông cân
\(\Rightarrow\widehat{HDA}=45^o\)
Xét \(ΔCED\&ΔCBA\) có :
\(\widehat{C}\) là góc chung
\(\widehat{D}=\widehat{A}=90^o\)
\(⇒ΔCED\simΔCBA( g.g)\)
\(\Rightarrow\dfrac{CD}{CA}=\dfrac{CE}{CB}\)
\(\Rightarrow\dfrac{CD}{CE}=\dfrac{CA}{CB}\)
Xét \(ΔCAD\&ΔCBE\):
\(\widehat{C}\) là góc chung
\(\dfrac{CD}{CE}=\dfrac{CA}{CB}\left(cmt\right)\)
\(\RightarrowΔCAD\simΔCBE\)
⇒\(\widehat{BEC}=\widehat{ADC}\)
Tacó \(\widehat{ADC}+\widehat{HDA}=180^o\)\(\widehat{BEC}+\widehat{BEA}=180^o\)
\(\Rightarrow\widehat{BEC}=\widehat{ADC}\)
\(\Rightarrow\widehat{HDE}=\widehat{BEA}\)
mà \(\widehat{HDE}=45^o\)°
\(\Rightarrow\widehat{BEA}=45^o\)
Xét \(ΔABE\)
\(\widehat{A}=90^o;\widehat{BEA}=45^o\)
\(\Rightarrow\Delta ABE\) vuông cân
\(\Rightarrow AB=AE\)
b) Ta có :
\(\widehat{AMB}=\widehat{AHB}=90^o\)
Nên tứ giác \(AMHB\) là tứ giác nội tiếp đường tròn đường kính \(AB\)
\(\Rightarrow\widehat{AHM}=\widehat{ABM}=45^o\)