Proof
Consider a triangle ABC in which $AC^2  = AB^2  + BC^2$.
We have to prove that ∠ B = 90°. 
	 



Construct a ∆ PQR right angled at Q such that PQ = AB and QR = BC Click Here.

Now, from ∆ PQR, we have :

$PR^2  =  PQ^2   +  QR^2$ (Pythagoras Theorem, as ∠ Q = 90°)

or, $PR^2  =  AB^2  + BC^2$     (By construction)......  (1)  
But  given that    $AC^2  =  AB^2  + BC^2$ ..........(2)
So, from (1) and  (2) ; AC =  PR  .............(3) 
Now, in ∆ABC and ∆ PQR,
	AB =  PQ   (By construction) 
	BC =  QR    (By construction)
	AC =  PR  [Proved in (3) above]

   So, ∆ ABC ≅ ∆ PQR      
   Therefore,    ∠ B =  ∠ Q        
But        ∠ Q =  90°
So,      ∠ B =  90°                                                  
Hence Proved

Example:
In Fig., ∠ ACB = 90° and CD ⊥ AB. 
	
Proof that: 
          ${BC^2}/{AC^2}={BD}/{AD}$
		  
Solution:
	∆ACD ~  ∆ ABC (Theorem   6.7) 
	So, ${AC}/{AB}={AD}/{AC}$
	or, $AC^2=AB.AD$ ...........(1)
	Similarly, ∆ BCD ~  ∆ BAC         (Theorem 6.7)
	So, ${BC}/{BA}={BD}/{BC}$
	or, $BC^2=BD.BA$ .............(2)
	Therefore, from (1) and (2),
	${BC^2}/{AC^2}={BD.BA}/{AB.AD}$
	So,${BC^2}/{AC^2}={BD}/{AD}$ hence Proved.
	

Example: A ladder is placed against a wall such that its foot is at a distance of 2.5 m from the
 wall and its top reaches a window 6 m above the ground. Find the length of the ladder. 

  Solution :
Let AB be the ladder and CA be the wall with the window at A (see Fig. 6.49).
Also,                               BC = 2.5 m and  CA = 6 m
From Pythagoras Theorem, we have: 
	$AB^2  =  BC^2  + CA^2$
		   = $(2.5)^2+ 6^2$
		   =42.25
So,    AB  = 6.5 
Thus, length of the ladder is 6.5 $m$.	


Example: In Fig., if AD ⊥ BC, prove that
       $AB^2  + CD^2  = BD^2  + AC^2$.


Solution:
			  From ∆ADC, we have
			  $AC^2  =  AD^2  + CD^2$ .........(1)
			  
			  From ∆ADB, we have
			  $AB^2  =  AD^2  + BD^2$ .............(2)
			  
			  Subtracting (1) from (2), we have
			  $AB^2  – AC^2  =  BD^2   –  CD^2$
			  or $AB^2  + CD^2  =  BD^2  + AC^2$
			  
 

 Example: BL and CM are medians of a triangle ABC right angled at A. Prove that
                $4 (BL^2  + CM^2) = 5 BC^2$.
 Solution:
 
Observe that the BL, CM, and BC , all are the hypotenuse of the 
triangles ∆ABL, ∆CMA and ∆ABC, so draw these triangles separately. Draw

		BL and CM are medians of the ∆ABC in which ∠A = 90° (see Fig.).
		From ∆ABC,
			$BC^2  =  AB^2  + AC^2 $
		From ∆ABL,
		     $BL^2  =  AL^2  + AB^2$
		Since L is the mid point of AC, therefore ${AL={AC}/2}$
		therefore, $BL^2  =  ({AC}/2)^2  + AB^2$
		           $BL^2  = {AC}^2/4  + AB^2$
			i.e    $4BL^2  = {AC}^2  + 4AB^2$ ...........(1)
			
		 From ∆ CMA,
		       $CM^2  =  AC^2  + AM^2$
			Since M is the mid point of AB, therefore ${AM={AB}/2}$
        Therefore, $CM^2  =  AC^2  + ({AB}/2)^2$ 
         i.e   $4CM^2  =  4AC^2  + {AB}^2$	 ..................(2)	
		 
		 Adding (1) and  (2)
		 $4BL^2+4CM^2  = {AC}^2  + 4AB^2 +4AC^2  + {AB}^2$
		 i.e $4(BL^2+CM^2)= 5({AC}^2+AB^2)$
		 Since $({AC}^2+AB^2)=BC^2$
		 Therefore $4(BL^2+CM^2)=5BC^2$ hence proved.

 

 Example:  O  is  any  point  inside  a rectangle ABCD (see Fig.). 
       Prove that  $OB^2  + OD^2  = OA^2  + OC^2$.
	
	
	Solution :
     Through O, draw PQ || BC so that P lies on AB and Q lies on DC.Draw
	 Now,       PQ ||  BC
	 Therefore,  PQ ⊥ AB  and PQ ⊥ DC (∠ B = 90° and ∠ C = 90°)
	So,  ∠ BPQ = 90°  and ∠CQP = 90°
	Therefore, BPQC and APQD are both rectangles.
	Now, from ∆ OPB,
	                $OB^2  =  BP^2   +  OP^2 $........(1)
					
	Similarly, from ∆OQD, 
					$OD^2  =  OQ^2   + DQ^2 $ ..........(2)
					
	From ∆ OQC, we have
				   $OC^2  =  OQ^2  + CQ^2 $ ............(3)
				   
	and from ∆ OAP, we have
					$OA^2  =  AP^2   +  OP^2$............(4)
	
	Adding (1) and (2),
	             $OB^2  + OD^2  =  BP^2  + OP^2  + OQ^2  + DQ^2$
				 Since  BP = CQ and DQ = AP
	            $OB^2  + OD^2=CQ^2  + OP^2  + OQ^2  + AP^2$
				             $=CQ^2  + OQ^2  + OP^2  + AP^2$
							 $=OC^2  + OA^2$
		Hence $OB^2  + OD^2  = OA^2  + OC^2$.