The blue triangle is similar to the green one and has the same orientation. The midpoints between corresponding vertices are connected to form a red triangle. Prove that it is also similar.

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## Solution

We span the blue and green triangle with vectors **a**, **b** and kR**a**, kR**b** respectively, where k is a scale factor and R is a counter-clockwise rotation over an angle β.

Now doing some simple vector algebra it is easy to see that the red triangle is spanned by the vectors (**a**+kR**a**)/2 and (**b**+kR**b**)/2. Let’s first calculate their lengths using the dot product. Using the fact that R**a·**R**a**=**a·a** and **a·**R**a**=R**a·a**=**a·a** cos(β), we find that the length of (**a**+kR**a**)/2 is K=√(1+k^{2}+2kcos(β))/2 times that of **a**. The calculation for **b** gives the same scaling factor K.

Now we take the mutual dot product. Using the fact that R**a·**R**b**=**a·b**=|**a**||**b**|cos(α) and **a·**R**b**=|**a**||**b**|cos(α+β) and R**a·b**=|**a**||**b**|cos(α-β) we get: (**a**+kR**a**)/2**·**(**b**+kR**b**)/2=K^{2}**a·b**.

We conclude that all three sides of the red triangle have the length of the corresponding blue side times K and hence the red triangle is similar.

Note that the area of the red triangle is K^{2} that of the blue one.