Consider right triangle ABC, right angled at B. Let BD be the perpendicular to the 
hypotenuse AC .
Proof 

	 

 
 In ∆ADB  and ∆ABC   
 ∠A =  ∠A
and ∠ADB =  ∠ ABC  
So, ∆ADB ~  ∆ ABC  ..........(1) 
Similarly,  ∆ BDC ~  ∆ ABC ........(2)   
So, from (1) and (2), triangles on both sides of the perpendicular BD are similar to the whole triangle ABC.
Also, since    ∆ADB ~  ∆ ABC
and       ∆ BDC ~  ∆ ABC
So,		∆ADB ~  ∆BDC    
 

Theorem  6.7  :  If  a  perpendicular  is  drawn  from the vertex of the right angle of a right triangle
 to the  hypotenuse  then  triangles  on  both  sides  of the perpendicular are similar to the whole triangle
 and  to  each  other.
 

Theorem  6.8  : In  a  right  triangle,  the  square  of  the  hypotenuse  is  equal  to  the sum  of  the  
squares  of  the  other  two  sides.

Proof : Consider a right triangle ABC right angled at B. 
We have to proof that $AC^2=AB^2+BC^2$
draw      BD ⊥ AC  as shown in the figure.
Now accordingly to Theorem 6.7, ∆ADB ~  ∆ ABC    
So, ${AD}/{AB}={AB}/{AC}$  (Sides are proportional)
or ${AD}.AC=AB^2$ ..............(1)
Also, accordingly to Theorem 6.7,∆ BDC ~  ∆ ABC   
So, ${CD}/{BC}={BC}/{AC}$

or $CD.AC=BC^2$ .............(2)

Adding (1) and (2),
$AD . AC + CD . AC =  AB^2  + BC^2 $
or, $AC (AD + CD) =  AB^2  + BC^2 $
Since AD+CD=AC;
therefore, $AC (AC) =  AB^2  + BC^2 $
or,   $AC^2 =  AB^2  + BC^2 $