Question

In the given figure, D is the midpoint of side BC and AE ⊥ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that  

(i) `b^2 = p^2 + ax + a^2/4` 

(ii) `c^2 = p^2 - ax + a^2/4`

(iii) `(b^2 + c^2) = 2p^2 + 1/2 a^2` 

(iv) `(b^2 - c^2) = 2ax` 

 

Answer 1

(i) In right-angled triangle AEC, applying Pythagoras theorem, we have: 

`AC^2=AE^2+EC^2` 

⇒ `B^2=h^2(x+a/2)^2=h^2+x^2+a^2/4+ax`.......(1) 

In right – angled triangle AED, we have: 

`AD^2=AE^2+ED^2` 

⇒ `p^2=h^2+x^2` .............(2) 

Therefore, from (i) and (ii), 

`b^2=p^2+ax+a^2/x`  

(ii) In right-angled triangle AEB, applying Pythagoras, we have: 

`AB^2=AE^2+EB^2` 

⇒ `c^2=h^2+(a/2-x)^2   (∵ BD=a/2 and BE=BD-x)` 

⇒ `C^2=h^2+x^2-a^2/4    (∵ h^2+x^2=p^2)` 

 ⇒`c^2=p^2-ax+a^2/x` 

(iii) Adding (i) and (ii), we get:  

⇒` b^2+c^2=p^2+ax+a^2/4+p^2-ax+a^2/4` 

`=2p^2+ax-ax+(a^2+a^2)/4` 

=`2p^2+a^2/2` 

(iv) Subtracting (ii) from (i), we get: 

`b^2-c^2=p^2+ax+a^2/4-(p^2-ax+a^2/4)` 

`=p^2-p^2+ax+ax+a^2/4-a^2/4` 

`=2ax`