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

Applying Pythagoras theorem for ΔACD, we obtain

AC2 = AD2 + DC2


AD2 = AC2 – DC2 (i)


Applying Pythagoras theorem in ΔABD, we obtain


AB2 = AD2 + DB2


AD2 = AB2 – DB2 (ii)


From (i) and (ii), we get


AC2 – DC2 = AB2 – DB2 (iii)


It is given that 3DC = DB


Therefore,


DC = and DB =


Putting these values in (iii), we get


AC2 2 = AB2 2


AC2 = AB2


16AC2 – BC2 = 16AB2 – 9BC2


16AB2 – 16AC2 = 8BC2


2AB2 = 2AC2 + BC2


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