In Figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC^2= AB^2+ BC^2 – 2 BC.BD.
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06/06/2021 12:50 pm
By applying Pythagoras Theorem in ∆ADB, we get,
AB^{2} = AD^{2} + DB^{2}
We can write it as;
⇒ AD^{2} = AB^{2} − DB^{2} ……………….. (i)
By applying Pythagoras Theorem in ∆ADC, we get,
AD^{2} + DC^{2} = AC^{2}
From equation (i),
AB^{2} − BD^{2} + DC^{2} = AC^{2}
AB^{2} − BD^{2} + (BC − BD)^{ 2} = AC^{2}
AC^{2} = AB^{2} − BD^{2} + BC^{2} + BD^{2} −2BC × BD
AC^{2 }= AB^{2} + BC^{2} − 2BC × BD
Hence, proved.
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