a) Ta có \(\widehat{ABE}+\widehat{BAH}=90^o=>\widehat{ABE}+\widehat{AFI}=90^o\)
\(=>\widehat{ABE}+\widehat{BFK}=90^o=>\widehat{CFK}+\widehat{IFH}=180^o-\widehat{AFI}-\widehat{BFK}=90^o\)
Do \(\widehat{FBE}=\widehat{BFK}=>FI\perp FK\)
b) \(AH=6;BC=8=>FI=\dfrac{1}{2}AH=\dfrac{1}{2}.6=3\)
\(BC=8=>FK=\dfrac{1}{2}BC=4\)
ΔFIK vuông tại F \(=>IK=\sqrt{FI^2+FK^2}=5\)