Two equilateral triangles sharing a vertex and three line segments. Prove that the red triangle is isosceles.

Scroll down for a solution to this problem.

## Solution

We provide a solution to the more general case of two similar isosceles triangles both having top angle α smaller than 180°. Construct vectors **a** and **b** along the sides as shown, R being a counterclockwise rotation over α. The vector connecting common vertex and midpoint is then (**a**+R**b**)/2, whereas the bottom vertices are connected by **b**-R**a**.

Now taking the dot product of those two, it must be 0 because of the right angle. So, (**a**+R**b**)**·**(**b**-R**a**)=0. This works out as **a·b**-R**a·**R**b**=**a·**R**a**–**b·**R**b**. Now the left side is 0, because rotations do not change the dot product. The right side equals (|**a**|^{2}-|**b**|^{2})cos(α).

So, either α=90° which is the situation in Meeting halfway, or the triangles are congruent.

## 4 replies on “The gate”

If this pdf file shows up, it is another, perhaps simpler, proof of the original, ungeneralized problem.

Here I try a jpeg file as requested.

One more way for the generalisation:

A mistake in my post (I moved B1 after writing the text):

The text in the picture above was made for B1 at the left of C.

A and B in the figure are on the same side of (B1 C1) and so C is on k1 as the angle(B1, C, C1) = alpha. – Sorry