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In `Delta ABC, AB= AC`. Side BC is produced to D. Prove that `(AD^(2)-AC^(2))=BD*CD`.

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In `Delta ABC, AB= AC`. Side BC is produced to D. Prove that `(AD^(2)-AC^(2))=BD*CD`.
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Draw `AE bot BC`. Then BE=CE
`AD^(2)=AE^(2)+DE^(2) and AC^(2)=AE^(2)+CE^(2)`
`rArr (AD^(2)-AC^(2))=DE^(2)-CE^(2)`
`=(DE+CE (DE-CE)`
`= (DE+BE)(DE-CE)[ :. CE=BE]`
`= BD*CD`
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