Question

Prove that in a parallelepiped the sum of the diagonals is equal to the sum of the squares on the edges.

Answer 1

Answer on Question #43544 – Math – Geometry

Prove that in a parallelepiped the sum of the diagonals is equal to the sum of the squares on the edges.

Proof.

A parallelepiped is a three-dimensional geometric solid with six faces that are parallelograms.

Parallelograms which is composed of a parallelepiped, called its faces, their side - ribs, and the vertices of parallelograms - tops the parallelepiped.

Based on the definition of a parallelepiped, we know that the sides are parallelograms, and hence in the proof will be guided with parallelogram properties.

We solve the auxiliary problem: we establish a relationship between the sides of parallelogram and its diagonals.

We proceed from the fact that the diagonal cross-section parallelepiped - parallelogram.

For parallelogram true each of the following statements:

- The opposing sides are equal in pairs;

- Opposite angles are equal;

- Diagonals intersect and bisect the intersection point;

- The opposing sides are parallel;

- The sum of angles adjacent to the one side, is $180{}^{\circ}$

- Each diagonal divides the parallelogram into two congruent triangles.

Consider the following parallelogram ABCD.

We have $\mathrm{AC} = \mathrm{d}_1, \mathrm{DB} = \mathrm{d}_2, \mathrm{AD} = \mathrm{BC} = \mathrm{a}, \mathrm{AB} = \mathrm{DC} = \mathrm{b}, \angle \mathrm{DAB} = \alpha$

Then $\angle ADC = 180{}^{\circ} - \alpha$

For triangle $\Delta$ DAB we note the law of cosines.

$$\mathrm {d} _ {2} ^ {2} = \mathrm {a} ^ {2} + \mathrm {b} ^ {2} - 2 \mathrm {a b} \cdot \cos (\alpha)$$

Then we consider the triangle $\Delta$ ADC and write for them the law of cosines.

$$\mathrm {d} _ {1} ^ {2} = \mathrm {a} ^ {2} + \mathrm {b} ^ {2} - 2 \mathrm {a b} \cdot \cos (1 8 0 {}^ {\circ} - \alpha) = \mathrm {a} ^ {2} + \mathrm {b} ^ {2} + 2 \mathrm {a b} \cdot \cos (\alpha)$$

Adding these equations, we obtain:

$$\mathrm {d} _ {1} ^ {2} + \mathrm {d} _ {2} ^ {2} = \mathrm {a} ^ {2} + \mathrm {b} ^ {2} + 2 \mathrm {a b} \cdot \cos (\alpha) + \mathrm {a} ^ {2} + \mathrm {b} ^ {2} - 2 \mathrm {a b} \cdot \cos (\alpha)$$

Simplify the obtained equation.

$$\mathrm {d} _ {1} ^ {2} + \mathrm {d} _ {2} ^ {2} = 2 \mathrm {a} ^ {2} + 2 \mathrm {b} ^ {2}$$

Consider the parallelepiped $AA_{1}DD_{1}CC_{1}BB_{1}$ to prove statement from the condition of the given task.

Let the edges of the parallelepiped are a, b, c.

For the plane $\mathrm{DD}_1\mathrm{B}_1\mathrm{B}$ write the formula that we determined of the sum of the diagonals.

$$\mathrm {D} _ {1} \mathrm {B} ^ {2} + \mathrm {D B} _ {1} ^ {2} = 2 \mathrm {D B} ^ {2} + 2 \mathrm {B B} _ {1} ^ {2}$$

Then we consider the plane $\mathrm{AA}_1\mathrm{C}_1\mathrm{C}$ write the formula that we determined of the sum of the diagonals.

$$\mathrm {A} _ {1} \mathrm {C} ^ {2} + \mathrm {A C} _ {1} ^ {2} = 2 \mathrm {A C} ^ {2} + 2 \mathrm {A A} _ {1} ^ {2}$$

Now we add obtained earlier equalities:

$$\mathrm {D} _ {1} \mathrm {B} ^ {2} + \mathrm {D B} _ {1} ^ {2} + \mathrm {A} _ {1} \mathrm {C} ^ {2} + \mathrm {A C} _ {1} ^ {2} = 2 \mathrm {D B} ^ {2} + 2 \mathrm {B B} _ {1} ^ {2} + 2 \mathrm {A C} ^ {2} + 2 \mathrm {A A} _ {1} ^ {2}$$

Now, to solve this equation we make the change of the sides.

Accordance with the terms of the problem, we noted the following.

$$\mathrm {D B} = \mathrm {d} _ {2} \text { and } \mathrm {A C} = \mathrm {d} _ {1}$$

We have already identified that $\mathrm{d}_1^2 + \mathrm{d}_2^2 = 2\mathrm{a}^2 + 2\mathrm{c}^2$ .

Now we can rewrite the equation.

$$\begin{array}{l} D _ {1} B ^ {2} + D B _ {1} ^ {2} + A _ {1} C ^ {2} + A C _ {1} ^ {2} = 2 D B ^ {2} + 2 B B _ {1} ^ {2} + 2 A C ^ {2} + 2 A A _ {1} ^ {2} \\ = 2 \left(\mathrm {D B} ^ {2} + \mathrm {A C} ^ {2}\right) + 2 \mathrm {B B} _ {1} ^ {2} + 2 \mathrm {A A} _ {1} ^ {2} \\ \end{array}$$

According to the condition of the task we also know that $\mathrm{BB}_1 = \mathrm{AA}_1 = \mathrm{b}$ .

So, we can substitute into the equation.

$$\mathrm {D} _ {1} \mathrm {B} ^ {2} + \mathrm {D B} _ {1} ^ {2} + \mathrm {A} _ {1} \mathrm {C} ^ {2} + \mathrm {A C} _ {1} ^ {2} = 2 (2 \mathrm {a} ^ {2} + 2 \mathrm {b} ^ {2}) + 2 \mathrm {b} ^ {2} + 2 \mathrm {b} ^ {2}$$

Simplify and combine like terms on the right side of the equation.

$$\mathrm {D} _ {1} \mathrm {B} ^ {2} + \mathrm {D B} _ {1} ^ {2} + \mathrm {A} _ {1} \mathrm {C} ^ {2} + \mathrm {A C} _ {1} ^ {2} = 2 (2 \mathrm {a} ^ {2} + 2 \mathrm {c} ^ {2}) + 2 \mathrm {b} ^ {2} + 2 \mathrm {b} ^ {2} = 4 \mathrm {a} ^ {2} + 4 \mathrm {c} ^ {2} + 4 \mathrm {b} ^ {2}$$

Thus we have the sum of squares of all the edges of the parallelepiped

Finally we can write the proof.

$$\mathrm {D} _ {1} \mathrm {B} ^ {2} + \mathrm {D B} _ {1} ^ {2} + \mathrm {A} _ {1} \mathrm {C} ^ {2} + \mathrm {A C} _ {1} ^ {2} = 4 \mathrm {a} ^ {2} + 4 \mathrm {c} ^ {2} + 4 \mathrm {b} ^ {2} = 4 (\mathrm {a} ^ {2} + \mathrm {c} ^ {2} + \mathrm {b} ^ {2}).$$

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