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For now, I will only post pictures of the theorems. Perhaps later I will add links that talk about the theorems in detail

Potema's theorem: Two squares created on the sides of a triangle $∆ABC$ And we set the point $M$ Which represents the middle $EF$ It will be constant no matter how you move point $C$ As long as $A,B$ are constant

Fensler-Hadwiger theorem We have two squares with a common vertex between them, so they will be the midpoints of the spaces between the opposite vertices, and the centers of the two squares will form the vertices of a third square

Two squares with a common vertex, the previous three properties will be fulfilled

If two squares have a common vertex, the two red lines will be perpendicular and equal

If we have two identical squares with one vertex at the center of the other, the common area between the two squares will be equal to a quarter of the area of ​​one square

In the previous figure there are two properties achieved: The first is that $AM=BM$ The second characteristic is that $PQ=2MN$

Two identical or different squares intersect and are enclosed between them $A,B,C,D$ The aforementioned equality will be achieved

If two identical or different squares have a common vertex, the blue area will be equal to the red area

If two squares have a common vertex, these three lines will meet at one point Moreover, the circles of the two squares pass through the same point, which represents the center of the spiral similarity

If two squares have parallel sides, these lines will meet at one point Furthermore itIf the two squares do not have parallel sides, then these lines will meet at one point. In the previous figure it would be: $\frac{b}{a}=\frac{\sqrt{5}+1}{2}=Φ$

In the previous picture it is: $X=\frac{5(A-B)}{8}$

We have two squares that have a common vertex, the point $F$ moves freely along the straight line $CD$ It will be points $A,C,Q$ Located on one straight line

Two squares in the plane In the general case, these two lines will be perpendicular

Here are some theorems that can be understood from pictures only without words:

For now, I will only post pictures of the theorems. Perhaps later I will add links that talk about the theorems in detail

Potema's theorem: Two squares created on the sides of a triangle $∆ABC$ And we set the point $M$ Which represents the middle $EF$ It will be constant no matter how you move point $C$ As long as $A,B$ are constant

Fensler-Hadwiger theorem We have two squares with a common vertex between them, so they will be the midpoints of the spaces between the opposite vertices, and the centers of the two squares will form the vertices of a third square

Two squares with a common vertex, the previous three properties will be fulfilled

If two squares have a common vertex, the two red lines will be perpendicular and equal

If we have two identical squares with one vertex at the center of the other, the common area between the two squares will be equal to a quarter of the area of ​​one square

In the previous figure there are two properties achieved: The first is that $AM=BM$ The second characteristic is that $PQ=2MN$

Two identical or different squares intersect and are enclosed between them $A,B,C,D$ The aforementioned equality will be achieved

If two identical or different squares have a common vertex, the blue area will be equal to the red area

If two squares have a common vertex, these three lines will meet at one point Moreover, the circles of the two squares pass through the same point, which represents the center of the spiral similarity

If two squares have parallel sides, these lines will meet at one point Furthermore itIf the two squares do not have parallel sides, then these lines will meet at one point. In the previous figure it would be: $\frac{b}{a}=\frac{\sqrt{5}+1}{2}=Φ$

In the previous picture it is: $X=\frac{5(A-B)}{8}$

We have two squares that have a common vertex, the point $F$ moves freely along the straight line $CD$ It will be points $A,C,Q$ Located on one straight line

Two squares in the plane In the general case, these two lines will be perpendicular

Here are some theorems that can be understood from pictures only without words:

For now, I will only post pictures of the theorems. Perhaps later I will add links that talk about the theorems in detail

Potema's theorem: Two squares created on the sides of a triangle $∆ABC$ And we set the point $M$ Which represents the middle $EF$ It will be constant no matter how you move point $C$ As long as $A,B$ are constant

Fensler-Hadwiger theorem We have two squares with a common vertex between them, so they will be the midpoints of the spaces between the opposite vertices, and the centers of the two squares will form the vertices of a third square

Two squares with a common vertex, the previous three properties will be fulfilled

If two squares have a common vertex, the two red lines will be perpendicular and equal

If we have two identical squares with one vertex at the center of the other, the common area between the two squares will be equal to a quarter of the area of ​​one square

In the previous figure there are two properties achieved: The first is that $AM=BM$ The second characteristic is that $PQ=2MN$

Two identical or different squares intersect and are enclosed between them $A,B,C,D$ The aforementioned equality will be achieved

If two identical or different squares have a common vertex, the blue area will be equal to the red area

If two squares have a common vertex, these three lines will meet at one point Moreover, the circles of the two squares pass through the same point, which represents the center of the spiral similarity

If two squares have parallel sides, these lines will meet at one point Furthermore itIf the two squares do not have parallel sides, then these lines will meet at one point. In the previous figure it would be: $\frac{b}{a}=\frac{\sqrt{5}+1}{2}=Φ$

In the previous picture it is: $X=\frac{5(A-B)}{8}$

We have two squares that have a common vertex, the point $F$ moves freely along the straight line $CD$ It will be points $A,C,Q$ Located on one straight line

Two squares in the plane In the general case, these two lines will be perpendicular

Here are some theorems that can be understood from pictures only without words:

For now, I will only post pictures of the theorems. Perhaps later I will add links that talk about the theorems in detail

Potema's theorem: Two squares created on the sides of a triangle $∆ABC$ And we set the point $M$ Which represents the middle $EF$ It will be constant no matter how you move point $C$ As long as $A,B$ are constant

Fensler-Hadwiger theorem We have two squares with a common vertex between them, so they will be the midpoints of the spaces between the opposite vertices, and the centers of the two squares will form the vertices of a third square

Two squares with a common vertex, the previous three properties will be fulfilled

If two squares have a common vertex, the two red lines will be perpendicular and equal

If we have two identical squares with one vertex at the center of the other, the common area between the two squares will be equal to a quarter of the area of ​​one square

In the previous figure there are two properties achieved: The first is that $AM=BM$ The second characteristic is that $PQ=2MN$

Two identical or different squares intersect and are enclosed between them $A,B,C,D$ The aforementioned equality will be achieved

If two identical or different squares have a common vertex, the blue area will be equal to the red area

If two squares have a common vertex, these three lines will meet at one point Moreover, the circles of the two squares pass through the same point, which represents the center of the spiral similarity

If two squares have parallel sides, these lines will meet at one point Furthermore itIf the two squares do not have parallel sides, then these lines will meet at one point. In the previous figure it would be: $\frac{b}{a}=\frac{\sqrt{5}+1}{2}=Φ$

In the previous picture it is: $X=\frac{5(A-B)}{8}$

We have two squares that have a common vertex, the point $F$ moves freely along the straight line $CD$ It will be points $A,C,Q$ Located on one straight line

Two squares in the plane In the general case, these two lines will be perpendicular

Here are some theorems that can be understood from pictures only without words:

For now, I will only post pictures of the theorems. Perhaps later I will add links that talk about the theorems in detail

Potema's theorem: Two squares created on the sides of a triangle $∆ABC$ And we set the point $M$ Which represents the middle $EF$ It will be constant no matter how you move point $C$ As long as $A,B$ are constant

Fensler-Hadwiger theorem We have two squares with a common vertex between them, so they will be the midpoints of the spaces between the opposite vertices, and the centers of the two squares will form the vertices of a third square

Two squares with a common vertex, the previous three properties will be fulfilled

If two squares have a common vertex, the two red lines will be perpendicular and equal

If we have two identical squares with one vertex at the center of the other, the common area between the two squares will be equal to a quarter of the area of ​​one square

In the previous figure there are two properties achieved: The first is that $AM=BM$ The second characteristic is that $PQ=2MN$

Two identical or different squares intersect and are enclosed between them $A,B,C,D$ The aforementioned equality will be achieved

If two identical or different squares have a common vertex, the blue area will be equal to the red area

If two squares have a common vertex, these three lines will meet at one point Moreover, the circles of the two squares pass through the same point, which represents the center of the spiral similarity

If two squares have parallel sides, these lines will meet at one point Furthermore itIf the two squares do not have parallel sides, then these lines will meet at one point. In the previous figure it would be: $\frac{b}{a}=\frac{\sqrt{5}+1}{2}=Φ$

In the previous picture it is: $X=\frac{5(A-B)}{8}$

We have two squares that have a common vertex, the point $F$ moves freely along the straight line $CD$ It will be points $A,C,Q$ Located on one straight line

Two squares in the plane In the general case, these two lines will be perpendicular

Here are some theorems that can be understood from pictures only without words: enter image description here

For now, I will only post pictures of the theorems. Perhaps later I will add links that talk about the theorems in detail

Potema's theorem: Two squares created on the sides of a triangle $∆ABC$ And we set the point $M$ Which represents the middle $EF$ It will be constant no matter how you move point $C$ As long as $A,B$ are constant

Fensler-Hadwiger theorem We have two squares with a common vertex between them, so they will be the midpoints of the spaces between the opposite vertices, and the centers of the two squares will form the vertices of a third square

Two squares with a common vertex, the previous three properties will be fulfilled

If two squares have a common vertex, the two red lines will be perpendicular and equal

If we have two identical squares with one vertex at the center of the other, the common area between the two squares will be equal to a quarter of the area of ​​one square

In the previous figure there are two properties achieved: The first is that $AM=BM$ The second characteristic is that $PQ=2MN$

Two identical or different squares intersect and are enclosed between them $A,B,C,D$ The aforementioned equality will be achieved

If two identical or different squares have a common vertex, the blue area will be equal to the red area

If two squares have a common vertex, these three lines will meet at one point Moreover, the circles of the two squares pass through the same point, which represents the center of the spiral similarity

If two squares have parallel sides, these lines will meet at one point Furthermore itIf the two squares do not have parallel sides, then these lines will meet at one point. In the previous figure it would be: $\frac{b}{a}=\frac{\sqrt{5}+1}{2}=Φ$

In the previous picture it is: $X=\frac{5(A-B)}{8}$

We have two squares that have a common vertex, the point $F$ moves freely along the straight line $CD$ It will be points $A,C,Q$ Located on one straight line

Two squares in the plane In the general case, these two lines will be perpendicular

Here are some theorems that can be understood from pictures only without words: