In triangle ABC, AE is perpendicular to BC and AD is the median, then prove that AB^2+BC^2=2(AD^2+BD^2)
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see diagram.
b² +c²
= be²+ec² + 2ae² using Pythagoras theorem.
= bd² +ed² - 2 bd ed + ed²+ dc2+2 ed dc +2 ae²
= 2 bd² + 2 ed² +2ae²
= 2 bd² + 2 AD²
as a² = 4 bd² and AD = p
b² +c²
= be²+ec² + 2ae² using Pythagoras theorem.
= bd² +ed² - 2 bd ed + ed²+ dc2+2 ed dc +2 ae²
= 2 bd² + 2 ed² +2ae²
= 2 bd² + 2 AD²
as a² = 4 bd² and AD = p
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