a/ Áp dụng định lý Pytago vào \(ΔABC\) vuông tại \(A\)
\(→BC=\sqrt{AB^2+AC^2}=\sqrt{625}=25\) (cm)
\(BD\) là đường phân giác \(\widehat{B}\)
\(→\dfrac{AD}{DC}=\dfrac{BA}{BC}\)
\(→\dfrac{AD}{DC}=\dfrac{15}{25}\) hay \(\dfrac{AD}{DC}=\dfrac{3}{5}\)
\(→\dfrac{AD}{3}=\dfrac{DC}{5}=\dfrac{AD+DC}{3+5}=\dfrac{20}{8}=\dfrac{5}{4}\)
\(→AD=\dfrac{15}{4}\)
\(\\\)
b/ Xét \(ΔABC\) và \(ΔHBA\):
\(\widehat{B}\): chung
\(\widehat{BAC}=\widehat{BHA}\)
\(→ΔABC\backsim ΔHBA(g-g)\)
\(→\dfrac{AB}{BC}=\dfrac{AH}{AB}\)
\(→AB^2=AH.BC\)
\(\\\)
c/ Xét \(ΔABC\) và \(ΔKDC\):
\(\widehat{C}\): chung
\(\widehat{BAC}=\widehat{DKC}\)
\(→ΔABC\backsim ΔKDC(g-g)\)
\(→\dfrac{CB}{CA}=\dfrac{CD}{CK}\)
\(→CB.CK=CA.CD\)
