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Prove that sagments joining the mid points of adjacent sides of a quadrilatral always form a parallelogram

Note by Aman Sharma
2 years ago

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Vector geometry is hardly ever used on Brilliant when, really, it should be used a whole lot more. Let's say that the four vertices of a quadrilateral are defined by vectors \(a,b,c,d\). Then the the vectors between adjacent midpoints are

\(\dfrac { 1 }{ 2 } \left( a+b \right) -\dfrac { 1 }{ 2 } \left( b+c \right) \)
\(\dfrac { 1 }{ 2 } \left( c+d \right) -\dfrac { 1 }{ 2 } \left( d+a \right) \)

that is

\(\dfrac { 1 }{ 2 } (a-c)\)
\(\dfrac { 1 }{ 2 } (c-a)\)

which means they're parallel. Do same for the other set. Michael Mendrin · 2 years ago

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@Michael Mendrin Sir is there any way to prove it by euclid's geometry Aman Sharma · 2 years ago

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@Aman Sharma Well, draw a diagonal from opposite corners of the quadrilateral. That forms \(2)\) triangles. Then it's easy to prove that the diagonal is parallel to the line through the midpoints of the sides of each of the \(2\) triangles. Hence, they're both parallel. Do the same for the other set. This is really the same thing as with the vector proof, but can be implemented using Euclid's geometry. Michael Mendrin · 2 years ago

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