prove apollonious theorem
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Let in a ΔABC, D is the midpoint of BC.Prove that:
AB2+AC2=2CD2+2AD2
Given that BD=DC and we construct E such that AE=EC⟹AC=EC2 and DE||AB⟹DE=AB2
For the ΔDEC we have DC2=DE2+EC2⟹4DC2=AC2+AB2
we have AB2+AC2=2CD2+2CD2(1)
In ΔADE we have, AD2=AE2+DE2⟹AE2+DC2−EC2⟹AD2=DC2
Hence 2CD2=2AD2
Thus, we have
AB2+AC2=2CD2+2AD2
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Ankitkumar200314:
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