# The perpendicular from A on side BC of aΔABC intersects BC at D such that DB = 3 CD(see Fig. 6.55). Prove that 2AB2 = 2AC2 + BC2

We have two right angled triangles now  ΔACD and ΔABD

Applying Pythagoras theorem for ΔACD, we obtain

AD2 = AC2 – DC2    ......................eq(i)

Applying Pythagoras theorem in ΔABD, we obtain

AD2 = AB2 – DB2  ........................eq(ii)

Now we can see from equation i and equation ii that LHS is same. Thus,

From (i) and (ii), we get

AC2 – DC2 = AB2 – DB2 (iii)

It is given that 3DC = DB

Therefore,
DC + DB = BC

DC + 3DC = BC
4 DC = BC ........................eq(iv)

and also,
DC = DB/3
putting this in eq (iii)
DB =
So,

DC = and DB =

Putting these values in (iii), we get

AC2  = AB2

16AC2 – BC2 = 16AB2 – 9BC2

16AB2 – 16AC2 = 8BC2

2AB2 = 2AC2 + BC2

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