In triangle ABC, Angle A = 90°, AC = AB and 'D' is a point on AB produced. Prove that
DC²-BD² = 2AB. A
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Answer:In ACD
A=90
H^2= B^2+P^2
DC^2=AD^2+AC^2
DC^2=AC^2+[AB+BD] ^2
DC^2=AC^2+AB^2+BD^2+2AB.BD
DC^2-BD^2=AC^2+AB^2+2AB.BD
DC^2-BD^2=AB^2+AB^2+2AB.AD [SINCE AC=AB]
DC^2-BD^2=2AB[AB+BD]
DC^2-BD^2=2AB.AD [SINCE AB+BD=AD]
H. P
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