a/ Xét \(ΔHAE\) và \(ΔHBD\):
\(\widehat{HEA}=\widehat{HDB}\) ( \(=90°\) )
\(\widehat{AHE}=\widehat{BHD}\) (đối đỉnh)
\(→ΔHAE\backsim ΔHBD(g-g)\)
b/ \(ΔHAE\backsim ΔHBD→\widehat{HBD}=\widehat{HAE}\) hay \(\widehat{EBK}=\widehat{DAC}\)
\(KE//AD\) (vì \(EK⊥BC,AD⊥BC\) )
\(→\widehat{DAC}=\widehat{CEK}\) (so le trong)
mà \(\widehat{DAC}=\widehat{EBK}\)
\(→\widehat{EBK}=\widehat{CEK}\)
Xét \(ΔEBK\) và \(ΔCEK\):
\(\widehat{EBK}=\widehat{CEK}\) (cmt)
\(\widehat{EKB}=\widehat{CKE}\) (\(=90°\) )
\(→ΔEBK\backsim ΔCEK(g-g)\)
\(→\dfrac{KE}{KB}=\dfrac{KC}{KE}\)
\(↔KE^2=KB.KC\)
c/ Xét \(ΔBEC\) và \(ΔADC\):
\(\widehat C\):chung
\(\widehat{BEC}=\widehat{ADC}\) (\(=90°\) )
\(→ΔBEC\backsim ΔADC(g-g)\)
\(→\dfrac{CE}{CB}=\dfrac{CD}{CA}\)
\(↔\dfrac{CE}{CD}=\dfrac{CB}{CA}\)
Xét \(ΔCED\) và \(ΔCBA\):
\(\widehat C\):chung
\(\dfrac{CE}{CD}=\dfrac{CB}{CA}\) (cmt)
\(→ΔCED\backsim ΔCBA(c-g-c)\)
Xét \(ΔCID\) và \(ΔCKE\):
\(\widehat C\) :chung
\(\widehat{CID}=\widehat{CKE}\) (\(=90°\) )
\(→ΔCID\backsim ΔCKE(g-g)\)
\(→\dfrac{CI}{CD}=\dfrac{CK}{CE}\)
\(↔\dfrac{CI}{CK}=\dfrac{CD}{CE}\)
Xét \(ΔCED\) và \(ΔCKI\):
\(\widehat C\):chung
\(\dfrac{CI}{CK}=\dfrac{CD}{CE}\) (cmt)
\(→ΔCED\backsim ΔCKI\) mà \(ΔCED\backsim ΔCBA\)
\(→ΔCKI\backsim ΔCBA\)
\(→\widehat{CKI}=\widehat{CBA}\) mà 2 góc ở vị trí đồng vị
\(→IK//AB\)