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Question
In figure AB = BC and AD is perpendicular to CD.
Prove that: AC2 = 2BC. DC.
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Solution
Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
We consider the ΔACD and applying Pythagoras theorem we get,
AC2 = AD2 + DC2
= ( AB2 - DB2 ) + ( DB + BC )2
= BC2 - DB2 + DB2 + BC2 + 2DB.BC ...( Given, AB = BC )
= 2BC2 + 2DB.BC
= 2BC( BC + DB )
= 2BC . DC
Hence proved.
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