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

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.

## Comments

Sort by:

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.

Log in to reply

Sir is there any way to prove it by euclid's geometry

Log in to reply

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