i) ABC is a triangle, in which AC > AB; this is only for the inference that the altitude

from A on to BC lies towards B, that is point 'E' lies between B and D; this is the most

important inference for proving the given

ii) D is the midpoint of BC

iii) AE perpendicular BC

Required to prove: AB^2=AD^2-BC*DE+1/4BC^2

PROOF: [for easy convenience, let me denote all squares by * and multiplication by 'x']

1) Since AE is perpendicular to BC, ABE is a right triangle with angle E = 90 deg.

2) Hence applying Pythagoras theorem we have, AB* = AE* + BE*

3) In similar way from the right triangle, ADE, AD* = AE* + DE*, => AE* = AD* - DE*

4) Substituting for AE* from step 3 in step 2, we have,

AB* = AD* - DE* + BE*

5) From the figure (Kindly make according to the description given in data),

BE = BD - DE

6) Substituting this in (4), AB* = AD* - DE* + (BD-DE)*

7) Expanding, AB* = AD* - DE* + BD* + DE* - 2 x BD x DE

8) Cancelling -DE* and +DE* and substituting BD = (1/2)BC

{Since, D is mid point of BC}

we get, AB* = AD* + (1/4)BC* - BC x DE

Thus it is proved that "AB^2=AD^2-BC*DE+(1/4)BC^2"

Now, BD = BC/2(since, D is midpoint)

Therefore,

AB2= AD2+ BD2-2BD*ED

=AD2+ (BC/2)2– 2(BC/2)*ED

= AD2+ (BC/2)2–2(BC/2)*ED

= AD2– BC*DE + ¼ BC2