a) Ta có: AB⊥AC(ΔABC vuông tại A)
HE⊥AC(gt)
Do đó: AB//HE(Định lí 1 về từ vuông góc tới song song)
b) Ta có: \(\widehat{B}=\widehat{HAC}\)(\(=90^0-\widehat{C}\))
\(\Leftrightarrow\widehat{HAC}=60^0\)
hay \(\widehat{HAE}=60^0\)
Ta có: ΔAHE vuông tại E(HE⊥AC)
\(\Rightarrow\widehat{HAE}+\widehat{AHE}=90^0\)(hai góc nhọn phụ nhau)
\(\Rightarrow\widehat{AHE}=90^0-\widehat{HAE}=90^0-60^0\)
hay \(\widehat{AHE}=30^0\)
Ta có: \(\widehat{HAC}+\widehat{BAH}=\widehat{BAC}\)(Tia AH nằm giữa hai tia AB,AC)
\(\Leftrightarrow60^0+\widehat{BAH}=90^0\)
\(\Leftrightarrow\widehat{BAH}=90^0-60^0\)
hay \(\widehat{BAH}=30^0\)
Vậy: \(\widehat{AHE}=30^0\); \(\widehat{BAH}=30^0\)