Q. 174.1( 8 Votes )

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

Answer :

(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.


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