Giải thích các bước giải:

a.Xét $\Delta ABC,\Delta HAB$ có:

Chung $\hat B$

$\widehat{BAC}=\widehat{AHB}(=90^o)$

$\to \Delta ABC\sim\Delta HBA(g.g)$

$\to \dfrac{AB}{HB}=\dfrac{BC}{AB}$

$\to AB^2=BH.BC$

b.Ta có: 

$BC=\sqrt{AB^2+AC^2}=\sqrt{2^2+(2\sqrt3)^2}=4$

$AH.BC=AB.AC(=2S_{ABC})$

$\to AH=\dfrac{AB.AC}{BC}=\sqrt3$

c.Ta có:

$y+z=(AB+AH+HB)+(AC+AH+HC)=AB+AC+(HB+HC)+2AH=AB+AC+BC+2AH$

$\to y+z=x+2AH$

$\to y+z-x=2AH$

$\to y+z+x=2x+2AH$

$\to (y+z-x)(y+z+x)=2AH\cdot (2x+2AH)$

$\to (y+z-x)(y+z+x)=4AH\cdot (x+AH)$

$\to (y+z-x)(y+z+x)=4(AH\cdot x+AH^2)$

Ta có: $\Delta AHB\sim\Delta CHA(g.g)$

$\to \dfrac{AH}{CH}=\dfrac{AB}{AC}$

$\to AH.AC=AB.HC$

Tương tự: $AH.AB=AC.HB$

$\to AH.AB+AH.AC =HB.AC+HC.AB$

$\to AH.AB+AH.AC+AH.BC=HB.AC+HC.AB+AH.BC$

$\to AH.(AB+AC+BC)=HB.AC+HC.AB+AH.BC$

$\to AH.x=HB.AC+HC.AB+AB.AC$

$\to AH\cdot x+AH^2=AH^2+HB.AC+HC.AB+AB.AC$

$\to AH\cdot x+AH^2=HB.HC+HB.AC+HC.AB+AB.AC$

$\to AH\cdot x+AH^2=(HB+AB)(HC+AC)$

$\to 2AH\cdot x+AH^2=AH\cdot x+(HB+AB)(HC+AC)$

$\to 2AH\cdot x+AH^2=AH(BC+AB+AC)+(HB+AB)(HC+AC)$

$\to 2AH\cdot x+AH^2=AH(HB+HC+AB+AC)+(HB+AB)(HC+AC)$

$\to 2AH\cdot x+AH^2=AH(HB+AB+HC+AC)+(HB+AB)(HC+AC)$

$\to 2AH\cdot x+2AH^2=AH^2+AH(HB+AB+HC+AC)+(HB+AB)(HC+AC)$

$\to 2(AH\cdot x+AH^2)=(AH+HB+AB)(AH+HC+AC)$

$\to 2(AH\cdot x+AH^2)=yz$

$\to 4(AH\cdot x+AH^2)=2yz$

$\to (y+z-x)(y+z+x)=2yz$

$\to (y+z)^2-x^2=2yz$

$\to (y+z)^2-2yz=x^2$

$\to y^2+z^2=x^2$