# 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.   Hence, proved.

(iv) Subtracting (iii) and (v),  Hence, proved.

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