Along similar lines

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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We span the blue and green triangle with vectors a, b and kRa, kRb 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+kRa)/2 and (b+kRb)/2. Let’s first calculate their lengths using the dot product. Using the fact that RRa=a·a and Ra=Ra·a=a·a cos(β), we find that the length of (a+kRa)/2 is K=√(1+k2+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 RRb=a·b=|a||b|cos(α) and Rb=|a||b|cos(α+β) and Ra·b=|a||b|cos(α-β) we get: (a+kRa)/2·(b+kRb)/2=K2a·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 K2 that of the blue one.

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