a: Xét ΔABC có \(BC^2=AB^2+AC^2\)
nên ΔABC vuông tại A
b: \(AH=AB^2:AC=\dfrac{15^2}{20}=11.25\left(cm\right)\)
HC=HA+AC=11,25+20=31,25(cm)
\(HB=\sqrt{31.25^2-25^2}=18.75\left(cm\right)\)
\(BK=\dfrac{BH^2}{BA}=\dfrac{18.75^2}{15}=23.4375\left(cm\right)\)
\(HK=\sqrt{23.4375^2-18.75^2}=14.0625\left(cm\right)\)