a)CM: \(tgAHB\sim tgAKC\left(g.g\right)\)
\(\Rightarrow\dfrac{AH}{AK}=\dfrac{AB}{AC}\)
Mà có góc B chung nên: \(tgAKH\sim tgACB\left(cgc\right)\)
=> \(\dfrac{AH}{AB}=\dfrac{KH}{BC}\)
=> CosA=\(\dfrac{KH}{BC}\)
=> KH=BC.cosA
b)ta có: KH=BC.CosA=\(\dfrac{BC}{2}\)
Trong tg vuông BKM có: \(KM=\dfrac{BC}{2}\)
Tt trong tg MHC có: MH=\(\dfrac{BC}{2}\)
=> TG KHM đều