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If the vectors emanting from one corner of the rectangel are called a and b then.

(a) + (b) = one diagonal

(a) + (-b) = the other diagonal

and |(a) + (b)| = |(a) + (-b)| (the absolute value of the diagonal's scalars are equal)

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Q: How do you prove that the diagonals of a rectangle are congruent using vectors?
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How do you prove the diagonals of an isosceles triangle congruent?

You can't because triangles do not have diagonals but an isosceles triangle has 2 equal sides


How do you prove that a shape is a rectangle?

Check to see that -- it has four sides -- its two diagonals are equal.


How can you prove that a rectangle is really a rectangle?

In a rectangle, all the opposite sides, and angles are congruent to each other. All 4 angles of the rectangle, are right angles.


How to Prove the diagonals of a rectangle bisect each other?

In rectangle ABCD, diagonals AC and BD meet at E. Angles BAC and DCA are alternate (since AB and DC are parallel) and are therefore equal. The same is true of angles ABD and CDB. Also, AB = DC, so that triangles ABE and CDE are congruent. Thus, |AE| = |EC| and |BE| = |ED|, that is, the point E bisects both AC and BD.QED


How do you prove that the diagonals in a rhombus are perpendicular to each other?

I will outline a way to prove it for you. I will also five a simple vector proof for those that have studied vectors. For the first proof, one can often cite some of these as known facts or refer to theorems in a text. 1. First show that a rhombus is a parallelogram 2. Next, using the above, show that diagonals of the rhombus divide it into 4 congruent triangles. 3. Last, use CPCTC and not that all 4 middle angles are congruent so that are 90 degrees. From this is is easy to say that the diagonals are perpendicular. Hints. to prove 1, use the fact that all 4 sides of the rhombus are congruent and then use SSS to find two congruent triangles. Then use CPCTC to show that the angles are the same and find a transversal. Look at same side interior angles cut by that transversal and say something about them being parallel. 2. Use SSS again and find 4 congruent triangles and look at the diagonals. I will help more by giving you another proof using vectors that is really much more straightforward. A rhombus is a quadrilateral with all sides having equal length. This means that if two vectors, a and b that form the corner of a rhombus, then the magnitude of a and b are equal The diagonals of the parallelogram are precisely a+b and a-b. Now look at the dot product of a+b and a-b and see that it is zero and remember that a dot product of zero means the vectors are perpendicular or orthogonal The first part is a pure synthetic geometry approach and if anyone need more help to finish that, just ask, The second part is a vector proof which is elegant because it is so simple.

Related questions

Prove that the diagonals of rectangle are equal?

prove any two adjacent triangles as congruent


What are necessary when proving that the diagonals of a rectangle are congruent?

A ruler or a compass would help or aternatively use Pythagoras' theorem to prove that the diagonals are of equal lengths


How do you prove the diagonals of an isosceles triangle congruent?

You can't because triangles do not have diagonals but an isosceles triangle has 2 equal sides


How do you prove that a shape is a rectangle?

Check to see that -- it has four sides -- its two diagonals are equal.


How can you prove that a rectangle is really a rectangle?

In a rectangle, all the opposite sides, and angles are congruent to each other. All 4 angles of the rectangle, are right angles.


What are the proofs that a quadrilateral ia s parallelogram?

There are 5 ways to prove a Quadrilateral is a Parallelogram. -Prove both pairs of opposite sides congruent -Prove both pairs of opposite sides parallel -Prove one pair of opposite sides both congruent and parallel -Prove both pairs of opposite angles are congruent -Prove that the diagonals bisect each other


Prove diagonals are equal in a rectangle?

Suppose ABCD is a rectangle.Consider the two triangles ABC and ABDAB = DC (opposite sides of a rectangle)BC is common to both trianglesand angle ABC = 90 deg = angle DCBTherefore, by SAS, the two triangles are congruent and so AC = BD.


How to Prove the diagonals of a rectangle bisect each other?

In rectangle ABCD, diagonals AC and BD meet at E. Angles BAC and DCA are alternate (since AB and DC are parallel) and are therefore equal. The same is true of angles ABD and CDB. Also, AB = DC, so that triangles ABE and CDE are congruent. Thus, |AE| = |EC| and |BE| = |ED|, that is, the point E bisects both AC and BD.QED


How do you prove that the diagonals in a rhombus are perpendicular to each other?

I will outline a way to prove it for you. I will also five a simple vector proof for those that have studied vectors. For the first proof, one can often cite some of these as known facts or refer to theorems in a text. 1. First show that a rhombus is a parallelogram 2. Next, using the above, show that diagonals of the rhombus divide it into 4 congruent triangles. 3. Last, use CPCTC and not that all 4 middle angles are congruent so that are 90 degrees. From this is is easy to say that the diagonals are perpendicular. Hints. to prove 1, use the fact that all 4 sides of the rhombus are congruent and then use SSS to find two congruent triangles. Then use CPCTC to show that the angles are the same and find a transversal. Look at same side interior angles cut by that transversal and say something about them being parallel. 2. Use SSS again and find 4 congruent triangles and look at the diagonals. I will help more by giving you another proof using vectors that is really much more straightforward. A rhombus is a quadrilateral with all sides having equal length. This means that if two vectors, a and b that form the corner of a rhombus, then the magnitude of a and b are equal The diagonals of the parallelogram are precisely a+b and a-b. Now look at the dot product of a+b and a-b and see that it is zero and remember that a dot product of zero means the vectors are perpendicular or orthogonal The first part is a pure synthetic geometry approach and if anyone need more help to finish that, just ask, The second part is a vector proof which is elegant because it is so simple.


How do you prove that the diagonals of an isosceles trapezoid are equal?

Let's draw the isosceles trapezoid ABCD, where AD ≅ BC, and mADC ≅ mBCD. If we draw the diagonals AC and BD of the trapezoid two congruent triangles are formed, ∆ ADC ≅ ∆ BDC (SAS Postulate: If two sides and the angle between them in one triangle are congruent to the corresponding parts in another triangle, then the triangles are congruent). Since these triangles are congruent, AC ≅ BD.


How do you prove In a circle or congruent circles congruent central angles have congruent arcs?

Chuck Norris can prove it


Prove that a rhombus has congruent diagonals?

Since the diagonals of a rhombus are perpendicular between them, then in one forth part of the rhombus they form a right triangle where hypotenuse is the side of the rhombus, the base and the height are one half part of its diagonals. Let's take a look at this right triangle.The base and the height lengths could be congruent if and only if the angles opposite to them have a measure of 45⁰, which is impossible to a rhombus because these angles have different measures as they are one half of the two adjacent angles of the rhombus (the diagonals of a rhombus bisect the vertex angles from where they are drawn), which also have different measures (their sum is 180⁰ ).Therefore, the diagonals of a rhombus are not congruent as their one half are not (the diagonals of a rhombus bisect each other).