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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