Prove that in any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side.

Solution :

Given : ABC is a triangle and AD is the median.

To Prove :

(i)  $${AB}^2$$ + $${AC}^2$$ = 2[$${AD}^2$$ + $${BD}^2$$]

(ii)  $${AB}^2$$ + $${AC}^2$$ = 2$${AD}^2$$ + 2$$[{1\over 2}BC]^2$$

Construction : Draw AE $$\perp$$ BC

Proof : In right angled triangle ABE, by Pythagoras theorem,

$${AB}^2$$ = $${BE}^2$$ + $${AE}^2$$                  ………(1)

In right angled triangle AEC, by Pythagoras theorem,

$${AC}^2$$ = $${EC}^2$$ + $${AE}^2$$                   ……….(2)

(i)    Adding (1) and (2), we get

$${AB}^2$$ + $${AC}^2$$ = 2$${AE}^2$$ + $${BE}^2$$ + $${EC}^2$$

= 2$${AE}^2$$ + $${(BD – ED)}^2$$ + $${(DC + ED)}^2$$

= 2$${AE}^2$$ + $${BD}^2$$ + $${ED}^2$$ + $${DC}^2$$ + $${ED}^2$$

= 2($${AE}^2$$ + $${ED}^2$$) + $${BD}^2$$ + $${DC}^2$$

Since in right triangle AED, $${AE}^2$$ + $${ED}^2$$ = $${AD}^2$$   and   BD = CD

= 2$${AD}^2$$ + $${BD}^2$$ + $${BD}^2$$

= 2[$${AD}^2$$ + $${BD}^2$$]

$$\therefore$$  BC = 2BD
$${AB}^2$$ + $${AC}^2$$ = 2$${AD}^2$$ + 2$$[{1\over 2}BC]^2$$