IX Proof of Heron’s formula

Let a, b, c are length of the sides and h is height to side of length c  of ∆  ABC.

We have S = (a + b + c)/2

So, 2s = a + b + c 

 Þ  2(s - a) = - a + b + c 
Þ  2(s - b) = a - b + c 
Þ  2(s - c) = a + b – c

Let p + q = c as indicated.

Then,   h² =  a² - p²       -------------(1)

 Also,   h² = b² - q ²       -------------- (ii)

From (i) and (ii)

  Þ  a² - p^{2     =}   b² -  q ²       

   Þ  q²   = - a² + p² + b²      

Since,   q = c - p  Þ q² = (c-p)² Þ q²  = c² + p² -2pc

Then, c² + p² -2pc   = - a² + p² + b²      

Þ - 2pc  =  - a² +b² – c² = - ( a² -b² + c²)

Þ   p = ( a² -b² + c²)/2c

Now, Put this value of p in equation (i)

h²    =   a² -  p²     

h²      =   ( a – p ) ( a + p )

h²   =    {a – ( a² -b² + c²)/2c }   {a + ( a² -b² + c²)/2c }

h²   =   {(2ac - a² + b² - c²)/2c}x{(2ac+ a² - b² + c²)/2c}

h²      = {(b² – (a - c)² }{(a + c)² – b²}/4c²

h²     = {(b – a + c) (b + a - c)}{(a + c + b)(a + c – b )

h² = { 2(s - a) x 2(s - c)  x 2s x2(s - b)}/4c²

h² = { 4 s (s - a) x (s - c) x(s - b)}/c²

h = 2/c  √ s (s - a) x (s - b) x(s - c) 

½ h c = √ s (s - a) x (s - b) x(s - c) 

Area of triangle = √ s (s - a) x (s - b) x(s - c)
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