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In ∆ ABC, AD ⊥ BC.

Prove that AC^{2} = AB^{2} +BC^{2} − 2BC x BD

In ∆ ABC (Figure 3), AD ⊥ BC.

Prove that AC^{2} = AB^{2} +BC^{2} − 2BC x BD

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#### Solution

Applying Pythagoras theorem in ΔADB, we obtain

AD^{2} + DB^{2} = AB^{2}

⇒ AD^{2} = AB^{2} − DB^{2 } .....(1)

Applying Pythagoras theorem in ΔADC, we obtain

AD^{2} + DC^{2} = AC^{2}

AB^{2} − BD^{2} + DC^{2} = AC^{2} ...[Using equation (1)]

AB^{2 }− BD^{2} + (BC − BD)^{2} = AC^{2}

AC^{2} = AB^{2} − BD^{2} + BC^{2} + BD^{2 }−2BC x BD

AC^{2 }= AB^{2 }+ BC^{2 }− 2BC x BD

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