a) Xét ΔAEB và ΔAFC ta có:
\(\widehat{A}\) chung
\(\widehat{AEB}=\widehat{AFC}=90^0\)
\(\Rightarrow\Delta AEB\)∼\(\Delta AFC\left(g.g\right)\)
\(b.Xét\) \(\Delta HFB\) \(và\) \(\Delta HEC\) \(ta\) \(có:\)
\(\widehat{BFH}=\widehat{HEC}=90^0\\ \widehat{FHB}=\widehat{EHC}\left(đ.đ\right)\)
\(\rightarrow\Delta HFB\)∼\(\Delta HEC\left(g.g\right)\)
\(\rightarrow\dfrac{HE}{HF}=\dfrac{HC}{HB}\\ \Rightarrow HE.HB=HF.HC\)
\(c.Xét\) \(\Delta AMD\) \(ta\) \(có:\)
\(AD//HF\left(DM\perp AB,FH\perp AB\right)\\ \rightarrow\dfrac{AF}{AH}=\dfrac{AH}{AD}\left(1\right)\)
\(Xét\) \(\Delta AND\) \(ta\) \(có:\)
\(HE//DM\left(HE\perp AC,DM\perp AC\right)\\ \rightarrow\dfrac{FA}{AM}=\dfrac{AH}{AD}\left(2\right)\)
\(Từ\left(1\right)và\left(2\right)\Rightarrow\dfrac{FA}{AM}=\dfrac{AE}{AN}\\ \Rightarrow EF//MN\)