Given a right triangle ABC, in wch BC is trisected on points D and E. Prove that 8AE2 = 3AC2 + 5AD2
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I'm really sorry I don't have paper to solve this question but I'll explain you the solution.
In this question using Pythagoras theorem first find all the values of the 3 hypotenuses that is AC, AE And AD. Since the base is trisected convert the bases of all the three in the form of BC or the smallest side.
Now the main part comes.
We know LHS=RHS
therefore LHS-RHS=0
so when you'll put the values everything will get cancelled and come out to be 0.
Hence Proved
In this question using Pythagoras theorem first find all the values of the 3 hypotenuses that is AC, AE And AD. Since the base is trisected convert the bases of all the three in the form of BC or the smallest side.
Now the main part comes.
We know LHS=RHS
therefore LHS-RHS=0
so when you'll put the values everything will get cancelled and come out to be 0.
Hence Proved
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