a) Xét \(ΔABH\) và \(ΔCBA\):
\(\widehat B:chung\)
\(\widehat{BHA}=\widehat{BAC}(=90°)\)
\(→ΔABH\backsim ΔCBA(g.g)\)
Xét \(ΔBAH\) và \(ΔACH\):
\(\widehat{HBA}=\widehat{HAC}\) (cùng phụ \(\widehat C\) )
\(\widehat{BHA}=\widehat{AHC}(=90°)\)
\(→ΔBAH\backsim ΔACH(g.g)\)
b) Xét \(ΔBAK\) và \(ΔBHM\):
\(\widehat{ABK}=\widehat{HBM}\) (\(BK\) hay \(BM\) là đường phân giác \(\widehat B\) )
\(\widehat{BAK}=\widehat{BHM}(=90°)\)
\(→ΔBAK\backsim ΔBHM(g.g)\)
\(→\dfrac{BA}{BK}=\dfrac{BH}{BM}\)
\(↔BA.BM=BH.BK\)
\(\dfrac{BA}{BK}=\dfrac{BH}{BM}\)
\(↔\dfrac{BA}{BH}=\dfrac{BK}{BM}\)
\(ΔABH\backsim ΔCBA\)
\(→\dfrac{BA}{BH}=\dfrac{BC}{BA}\) mà \(\dfrac{BA}{BH}=\dfrac{BK}{BM}\)
\(→\dfrac{BC}{BA}=\dfrac{BK}{BM}\)
\(↔BA.BK=BC.BM\)
c) \(DK⊥HC,AH⊥HC\)
\(→DK//AH\)
\(→\dfrac{DH}{DC}=\dfrac{KA}{KC}\) (Định lý Talet)
\(BK\) là đường phân giác \(\widehat B\)
\(→\dfrac{KA}{KC}=\dfrac{BA}{BC}\) mà \(\dfrac{DH}{DC}=\dfrac{KA}{KC}\)
\(→\dfrac{BA}{BC}=\dfrac{DH}{DC}\)
\(↔\dfrac{BA}{DH}=\dfrac{BC}{DC}\)
