In figure, ∠B of ∆ABC is an acute angle and AD ⊥ BC, prove that AC2 = AB2 + BC2 – 2BC × BD
Given: A ∆ABC in which ∠B is an acute angle and AD ⊥ BC.
To Prove: AC2 = AB2 + BC2 – 2BC × BD.
Proof: Since ∆ADB is a right triangle right-angled at D. So, by Pythagoras theorem, we have
AB2 = AD2 + BD2 ….(i)
Again ∆ADC is a right triangle right angled at D.
So, by Pythagoras theorem, we have
AC2 = AD2 + DC2
⇒ AC2 = AD2 + (BC – BD)2
⇒ AC2 = AD2 + (BC2 + BD2 – 2BC • BD)
⇒ AC2 = (AD2 + BD2 ) + BC2 – 2BC • BD
⇒ AC2 = AB2 + BC2 – 2BC • BD [Using (i)]
Hence, AC2 = AB2 + BC2 – 2BC • BD
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