MHB Proving Orthogonal Projection of Triangle V, v'_{1}

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The discussion focuses on proving the equation \( v_{1}=V \cos(\psi)+v'_{1} \cos(\theta - \psi) \) related to the orthogonal projections of triangle components. It is established that \( v_{1} \) represents the sum of the orthogonal projections of \( V \) and \( v'_{1} \) onto \( v_{1} \). Participants express difficulty in visualizing and calculating these projections. Clarification on the geometric interpretation and mathematical steps for deriving the projections is sought. Understanding these projections is crucial for validating the given expression.
rannasquaer
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Given the triangle above where $$V < v'_{1}$$, prove that the \[ v_{1}=V \cos(\psi)+v'_{1} \cos(\theta - \psi) \]

It is said that $$v_{1}$$ is equal to the sum of the orthogonal projections on $$v_{1}$$ of $$V$$ and of $$v'_{1}$$ and that is precisely the expression that show. But I couldn't see how to make the projection and the calculations.
 
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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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