Question

ABC is a right triangle, right angled at A and D is mid point of AB. Prove that, BC²=CD²+3BD²

Answer 1

here is the answer....

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Answer 2

In triangle ABC
By pythagoras theorem
BC2=AB2+AC2....(1)

In triangle ADC
CD2=AC2+AD2
AC2=CD2-AD2

from (1)  
BC2=AB2+CD2-AD2
BC2=(AD+BD)2+CD2-AD2
BC2=AD2+2AD.BD+BD2+CD2-AD2
BC2=2AD.BD+BD2+CD2
since D is a midpoint of AB
AD=BD
BC2=2BD2+BD2+CD2
BC2=3BD2+CD2