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Question

In an isosceles triangle ABC with AB=AC, BD is perpendicular from B to side AC. Prove that BD2CD2=2CD.AD
283175_34497ace48dd4eb58cb3fb625af7ac45.png

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Solution

Since ADB is right-angled at D,

AB2=AD2+BD2

AC2=AD2+BD2 (AB=AC)

(AD+CD)2=AD2+BD2

AD2+CD2+2AD.CD=AD2+BD2

BD2CD2=2AD.CD

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