# prove it!!!!!!!!!!!!!!(4)

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

Note by Aman Sharma
3 years, 8 months ago

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## Comments

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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.

- 3 years, 8 months ago

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

- 3 years, 8 months ago

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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.

- 3 years, 8 months ago

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