goformit100
  • goformit100
Given a quadrilateral ABCD such that AB^2+CD^2=BC^2+AD^2, prove that AC⊥BD.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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goformit100
  • goformit100
@oldrin.bataku @shivamgautam
goformit100
  • goformit100
@vikrantg4
mathslover
  • mathslover
|dw:1371382939945:dw|

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mathslover
  • mathslover
|dw:1371384170597:dw|
anonymous
  • anonymous
I'll do it with analytic geometry: place the vertices at D=(0,0),C=(x,0),A=(a,b),B=(α,β) Then we get that AB^2+CD^2=(a−α)^2+(b−β)^2+x^2=b^2+(α−x)^2+a^2+b^2=BC^2+AD^2 implies, aα+bβ=xα or,αβ=b/x−a now slope of AC is b/a-x also, slope of BD is β/α slope os AB*CD gives us the required condition. Good problem though.
mathslover
  • mathslover
\(\mathsf{x_1^2 + y_1 ^2 + (x_2 - x_3)^2 + (y_2 - y_3)^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + x_3^2 + y_3 ^2 }\) \(\mathsf{-2x_2 x_3 -2y_2 y_3 = -2x_2 x_1 -2 y_2 y_1 \\ \implies 2x_2 x_1 - 2x_2 x_3 = 2y_2 y_3 - 2y_2 y_1 \\ \implies 2x_2 ( x_1 - x_3 ) = 2y_2 ( y_3 - y_1) \\ \implies \cfrac{x_2 }{ y_2} = \cfrac{y_3 - y_1}{x_1 - x_3} \\ \implies \cfrac{y_3 - y_1 }{ x_3 - x_1 } \times \cfrac{y_2 }{ x_2} = -1 }\\ \textbf{If slope of two lines is equal to -1 then both are perpendicular to }\\ \textbf{each other} \)
mathslover
  • mathslover
\(\cfrac{y_3 - y_1 }{ x_3 - x_1 } = \textbf{Slope of line BD} \) and \(\cfrac{y_2 }{ x_2} = \textbf{Slope of line AC} \) And by the above work it is clear that, \(\cfrac{y_3 - y_1}{x_3 - x_1} \times \cfrac{y_2 }{ x_2} = -1\) that is : Slope of line BD * Slope of line AC = -1 . Thus \(AC \perp BD\) \(\ast \textbf{Proved} \ast \)

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