ABC is an equilateral triangle with AD perpendicular to BC , prove tha

1.	ABC is an equilateral triangle with AD perpendicular to BC , prove that AD² = 3DC².
Answer» Here given triangle ABC must be an equilateral triangle and we have to prove AD2\xa0= 3DC2Solution:Given:\xa0∆ABC is an equilateral, AD ⊥ BC.\xa0\xa0In right ∆ABD,AB2\xa0= BD2\xa0+ AD2⇒ BC2\xa0= BD2\xa0+ AD2 (AB = BC)⇒ (2BD)2\xa0= BD2\xa0+ AD2 \xa0[From (1)]⇒ 4BD2\xa0= BD2\xa0+ AD2⇒ AD2\xa0= 3BD2\xa0= 3DC2 [As BD = DC]Hence proved

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