Q. 144.2( 234 Votes )

# The perpendicular from A on side BC of aΔABC intersects BC at D such that DB = 3 CD(see Fig. 6.55). Prove that 2AB^{2} = 2AC^{2} + BC^{2}

Answer :

We have two right angled triangles now ΔACD and ΔABD

Applying Pythagoras theorem for ΔACD, we obtain

AC^{2} = AD^{2} + DC^{2}

AD^{2} = AC^{2} – DC^{2} ......................eq(i)

Applying Pythagoras theorem in ΔABD, we obtain

AB^{2} = AD^{2} + DB^{2}

AD^{2} = AB^{2} – DB^{2} ........................eq(ii)

From (i) and (ii), we get

AC^{2} – DC^{2} = AB^{2} – DB^{2} (iii)

It is given that 3DC = DB

Therefore,

DC + DB = 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

AC^{2} – = AB^{2} – ^{}

16AC^{2} – BC^{2} = 16AB^{2} – 9BC^{2}

16AB^{2} – 16AC^{2} = 8BC^{2}

2AB^{2} = 2AC^{2} + BC^{2}

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