# How to prove area of the square and parallelogram the same?

Please show me how do you find length of BH or AI (in general), I want both specific proof for this case and in a general case.

My purpose is to try to prove that the area of the square is the same as the area of the parallelogram given two parallel line L1 and L2 and the square and the parallelogram share the same base AB.

• The easiest proof goes like this: This is a direct consequence of Cavalieri's principle. Q.E.D. – user137731 Sep 30 '16 at 22:20
• Hint: for a point $C$ on $L_2$ the area of $\triangle CAB$ is... P.S. The linear-algebra tag doesn't apply. – dxiv Sep 30 '16 at 22:20
• @ dxiv the area can be found through the determinant of the vector AB and AC. But I am trying to understand this geometrically. That's why the linear algebra tag is relevant in my opinion. – Tmm Sep 30 '16 at 22:37
• $AB$ times the distance between the parallel lines is twice the area of $\triangle CAB$ (why?). Then... – dxiv Sep 30 '16 at 22:54
• @ dxiv I just figured out. First, answer your question why △CAB is half of AB times the distance between the parallel lines. △CAB is half of the parallelogram so its area is half of the parallelogram. Also, the rectangle ABHI has the same area as the parallelogram is because the parallelogram is a transformation of the rectangle and transformation does not effect the determinant. Geometrically speaking, the area of the rectangle is Base ||AB|| times Height(the distance between the parallel lines). I used vector projection to find ||BH||, and verify the areas of ABHI & ABDC are the same. – Tmm Sep 30 '16 at 23:41

The area of any parallelogram is base times the height and in your case it is: $AB.BH$ which is the same as the area of the square.