Question 14:
The perpendicular from A on side BC of a ;ABC intersect BC at D such that DB = 3 CD. Prove that 2 AB2 =2 AC2 + BC2
Answer:
Applying Pythagoras theorem for ΔACD, we obtain
AC2 = AD2 + DC2
AD2 = AC2 – DC2 (1)
Applying Pythagoras theorem in ∆ABD, we obtain
AB2 = AD2 + DB2
AD2 = AB2 – DB2 (2)
From equation (1) and (2), we obtain
AC2 – DC2 = AB2 – DB2 (3)
It is given that 3DC = DB
∴DC= BC/4 and DB = 3BC/4
Putting these values in equation (3), we obtain
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