Xét \(\Delta APE\) và \(\Delta APH\) có :
PE = PH (gt)
PA : cạnh chung (gt)
\(\widehat{APE}=\widehat{APH}\left(=90^0\right)\)
\(\Rightarrow\Delta APE=\Delta APH\) (c . g . c)
\(\Rightarrow\widehat{EAP}=\widehat{HAP}\)
Xét \(\Delta AQF\) và \(\Delta AQH\) có :
AQ : cạnh chung
QH = QF (gt)
\(\widehat{AQH}=\widehat{AQF}\left(=90^0\right)\)
\(\Rightarrow\Delta AQH=\Delta AQF\) (c . g . c)
\(\Rightarrow\widehat{HAQ}=\widehat{FAQ}\)
Ta có : \(\widehat{QAH}+\widehat{PAH}=90^0\)
\(\Rightarrow\widehat{EAP}+\widehat{FAQ}=90^0\)
Mà \(\widehat{EAF}=\widehat{EAP}+\widehat{PAQ}+\widehat{FAQ}\)
\(=\widehat{EAP}+\widehat{FAQ}+\widehat{PAQ}\) \(=90^0+90^0=180^0\) \(\Rightarrow\) 3 điểm E,A,F thẳng hàng
a) Xét \(\Delta APE,\Delta APH\) có :
\(PE=PH\left(gt\right)\)
\(\widehat{APE}=\widehat{APH}\left(=90^{^O}\right)\)
\(AP:Chung\)
=> \(\Delta APE=\Delta APH\) (2 cạnh góc vuông)
Xét \(\Delta AQH,\Delta AQF\) có :
\(HQ=FQ\left(gt\right)\)
\(\widehat{AQH}=\widehat{AQF}\left(=90^o\right)\)
\(AQ:Chung\)
=> \(\Delta AQH=\Delta AQF\) (2 cạnh góc vuông)
b) Ta có : \(\widehat{PAH}+\widehat{QAH}=90^o\)
=> \(\widehat{EAP}+\widehat{FAQ}=90^o\)
Ta có : \(\widehat{EAP}+\widehat{PAH}+\widehat{QAH}+\widehat{FAQ}=180^o\)
Do đó: A,E,F thẳng hàng.
