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9AD^2=7AB^2

Answer» In an equilateral ABC, D is a point on side BC such that BD equals 1 third BC. Prove that 9AD2 = 7AB2. Given : An equilateral triangle ABC such that To Prove : 9AD2 = 7ABGiven : An equilateral triangle ABC such that BD equals 1 third BC.To Prove : 9AD2 = 7AB2Const: Draw AE ⊥ BC.Proof: In right triangles ABE and ACE, we haveAE = AE [common]∠AEB = ∠AEC [90°]and AB = AC[∆ABC is an equilateral]Therefore, by using RHS congruent condition space space increment ABE approximately equal to increment ACErightwards double arrow BE = CE [by CPCT]rightwards double arrow BE equals CE equals 1 half BCIn right triangle ADE, we have space AD squared space equals space AE squared plus DE squaredrightwards double arrow AD squared space equals space AE squared plus left parenthesis BE minus BD right parenthesis squaredrightwards double arrow space space space AD squared equals AE squared plus BE squared plus BD squared minus 2 BD. BErightwards double arrow AD squared space equals space left parenthesis AE squared plus BE squared right parenthesis plus BD squared minus 2 BD. BE open square brackets because space space BD equals 1 third BC space and space BE space equals space CE space equals space 1 half BC close square bracketsrightwards double arrow space space space AD squared space equals space AB squared plus open parentheses 1 third BC close parentheses squared minus 2 cross times 1 third BC cross times 1 half BCrightwards double arrow space space space AD squared space equals space AB squared plus 1 over 9 BC squared minus BC squared over 3rightwards double arrow AD squared equals fraction numerator 9 AB squared plus BC squared minus 3 BC squared over denominator 9 end fractionrightwards double arrow 9 AD squared space equals space 9 AB squared minus 2 BC squaredBut AB = BC = CArightwards double arrow 
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rightwards double arrow 
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In an equilateral ABC, D is a point on side BC such that\xa0\xa0Prove that\xa09AD2\xa0= 7AB2.\xa0
Given : An equilateral triangle ABC such that\xa0To Prove : 9AD2\xa0= 7AB2Const: Draw AE ⊥ BC.Proof: In right triangles ABE and ACE, we haveAE = AE [common]∠AEB = ∠AEC [90°]and AB = AC[∆ABC is an equilateral]Therefore, by using RHS congruent condition BE = CE [by CPCT]In right triangle ADE, we haveBut AB = BC = CA\xa0


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