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Question

In an equilateral ΔABC,ADBC. prove that AD2=3BD2

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Solution


We have, ΔABC is an equilateral Δ and AD ⊥ BC
In ΔADB and ΔADC
angleADB = angleADC { Each 90}
AB = AC {Given}
AD = AD {Common}
Then, ΔADB approximately equal to ΔADC [ By RHS condition]

BD = CD = BC2 --- (i) Corresponding parts of similar triangles
In, ΔABD, by Pythagoras theorem
A B squared equals A D squared plus B D squared B C squared equals A D squared plus B D squared space space space space space left square bracket A B space equals space A C space g i v e n right square bracket left parenthesis 2 B D right parenthesis squared equals A D squared plus B D squared space space space space space left square bracket F r o m space left parenthesis i right parenthesis right square bracket 4 B D squared equals A D squared plus B D squared space 3 B D squared equals A D squared space
Hence proved


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