Asked by Evelina
I need to prove parallelogram law using the law of cosines.
2AB^2 + 2BC^2 = AC^2 + BD^2
Please, help me to do this.
2AB^2 + 2BC^2 = AC^2 + BD^2
Please, help me to do this.
Answers
Answered by
Steve
Let
θ = angle DAB
Ø = angle ABC
By the law of cosines,
BD^2 = AD^2 + AB^2 - 2 AD AB cosθ
AC^2 = BC^2 + AB^2 - 2 BC AB cosØ
Now, since AD = BC,
BD^2 + AC^2 = 2AD^2 + 2AC^2 - 2 AD AB (cosθ + cosØ)
Since θ+Ø = 180°
cosθ = -cosØ
'nuff said.
θ = angle DAB
Ø = angle ABC
By the law of cosines,
BD^2 = AD^2 + AB^2 - 2 AD AB cosθ
AC^2 = BC^2 + AB^2 - 2 BC AB cosØ
Now, since AD = BC,
BD^2 + AC^2 = 2AD^2 + 2AC^2 - 2 AD AB (cosθ + cosØ)
Since θ+Ø = 180°
cosθ = -cosØ
'nuff said.
Answered by
Steve
Oops. In the long line, on the right side, AC should read AB
Answered by
Evelina
ehh, still don't get it.
Answered by
Steve
where do you get stuck? I'll try to clarify. Draw a diagram to follow along.
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