a,
$\Delta$ ABC, $\widehat{A}=90^o, AH\bot BC$:
$AB=\sqrt{BC^2-AC^2}=9cm$
$\Rightarrow AH=\dfrac{AB.AC}{BC}=7,2cm$
$HB=\sqrt{AB^2-AH^2}=5,4cm$
$\Rightarrow HC=15-5,4=9,6cm$
b,
$\Delta$ AHB, $\widehat{AHB}=90^o, HE\bot AB$:
$AH^2=AE.AB$ (1)
$\Delta$ AHC, $\widehat{AHC}=90^o$, $HF\bot AC$:
$AH^2=AF.AC$ (2)
(1)(2)$\Rightarrow AE.AB=AF.AC$
c,
Tứ giác HEAF có $\widehat{HEA}=\widehat{EAF}=\widehat{HFA}=90^o$ nên là hình chữ nhật.
$\Rightarrow \widehat{EHF}=90^o$
Theo Pytago, $HE^2+HF^2=EF^2$
Mà $AH=EF\Rightarrow HE^2+HF^2=AH^2$
$\Delta$ ABC có $AH^2=BH.HC$
Vậy $HE^2+HF^2=BH.HC$