\(CM:tgAHB\sim tgAKC\left(gg\right)\)
\(\Rightarrow\dfrac{AH}{AB}=\dfrac{AK}{AC}\)
mà B chung => \(tgAKH\sim tgACB\)
=>\(\dfrac{AH}{AB}=\dfrac{KH}{BC}\)
=> KH=cosA.BC
b) Ta có: KH=CosA.BC=\(\dfrac{BC}{2}\)
trong tgMHC có: MH=\(\dfrac{BC}{2}\)
CM tt với MK
=> tg đều