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

@Aman Sharma
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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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TopNewestVector 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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– Aman Sharma · 2 years ago

Sir is there any way to prove it by euclid's geometryLog in to reply

– Michael Mendrin · 2 years ago

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.Log in to reply