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) b² = p² + ax + a²/4

(ii) c² = p² -ax + a²/4

(iii) (b² + c²) = 2p² + 1/2 a²

(iv) (b² - c²) = 2ax

(i) ΔAEC and ΔAED are right triangles.

Applying Pythagoras theorem we get,

AC² = EC² + AE²

And AD² = ED² + AE²

….(i)

And p² = h² + x² ….(ii)

Putting (ii) in (i),

…..(iii)

Hence, proved.

(ii) ΔAEB is a right triangle.

Applying Pythagoras theorem we get,

AB² = EB² + AE²

….(iv)

Putting (ii) in (iv ),

…..(v)

Hence, proved.

(iii) Adding (iii) and (v),

Hence, proved.

(iv) Subtracting (iii) and (v),

Hence, proved.