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Ah, what a lovely question! To prove that the diagonals of a square are perpendicular, you simply need to show that the opposite sides are equal in length and that the diagonals bisect each other at a 90-degree angle. It's like painting a beautiful landscape, each part working together to create a harmonious whole. Just follow the gentle flow of geometry, and you'll see the beauty of the square unfold before your eyes.

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BobBot

4mo ago

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How do you prove that the diagonals of a square are equal and perpendicular to each other?

To prove that the diagonals of a square are equal and perpendicular, start by noting that a square has four equal sides and four right angles. By using the distance formula, calculate the lengths of the diagonals; since both diagonals connect opposite vertices, they will have the same length, confirming they are equal. To show they are perpendicular, observe that the slopes of the diagonals are negative reciprocals of each other, as they intersect at right angles, demonstrating that the diagonals are perpendicular. Thus, the properties of the square ensure that both conditions are satisfied.


Why are the diagonals of a square perpendicular bisectors of each other?

You could prove this by congruent triangles, but here are two simpler arguments: --------------- Since a square is a rhombus, and the diagonals of a rhombus are perpendicular bisectors of each other, then the diagonals of a square must be perpendicular bisectors of each other -------------------- A square has four-fold rotational symmetry - as you rotate it around the point where the diagonals cross, there are four positions in which it looks the same. This means that the four angles at the centre must be equal. They will each measure 360/4 = 90 degrees, so the diagonals are perpendicular. Also. the four segments joining the centre to a vertex are all equal, so the diagonals bisect each other.


In coordinate geometry How do you prove square by using section formula?

To prove that a quadrilateral is a square using the section formula, calculate the midpoints of the diagonals. If the midpoints of both diagonals are the same, the diagonals bisect each other, indicating a parallelogram. Next, verify that all sides are equal by calculating the lengths of each side using the distance formula. Lastly, confirm that the diagonals are equal in length and perpendicular, which is characteristic of a square.


Prove that if the diagonal of a parallelogram does not bisect the angles through the vertices to which the diagonal is drawn the parallelogram is not a rhombus?

Suppose that the parallelogram is a rhombus (a parallelogram with equal sides). If we draw the diagonals, isosceles triangles are formed (where the median is also an angle bisector and perpendicular to the base). Since the diagonals of a parallelogram bisect each other, and the diagonals don't bisect the vertex angles where they are drawn, then the parallelogram is not a rhombus.


Prove that the diagonals of rectangle are equal?

prove any two adjacent triangles as congruent


How do you prove that the diagonals of a square are equal and bisect each other at right angles?

it would produce two right angle triangleImproved Answer:-Measure them and use a protractor which will result in equal measures of 4 by 90 degrees angles


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


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).


Show that if diagonals of a quadrilateral bisects each other then it is a rhombus?

This cannot be proven, because it is not generally true. If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. And conversely, the diagonals of any parallelogram bisect each other. However not every parallelogram is a rhombus.However, if the diagonals are perpendicular bisectors, then we have a rhombus.Consider quadrilateral ABCD, with diagonals intersecting at X, whereAC and BD are perpendicular;AX=XC;BX=XD.Then angles AXB, BXC, CXD, DXA are all right angles and are congruent.By the ASA theorem, triangles AXB, BXC, CXD and DXA are all congruent.This means that AB=BC=CD=DA.Since the sides of the quadrilateral ABCD are congruent, it is a rhombus.


How do you prove the diagonal lines in a parallelogram is not a rhombus?

Because the diagonals of a rhombus intersect each other at 90 degrees whereas in a parallelogram they don't


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


What is the difference between a rhombus and a square?

The difference is a rhombus is made up of 2 acute and 2 obtuse angles and a square is made up of 4 right angles. On a square, both of the diagonals equal each other, but on a rhombus the diagonals can equal each other, but it's possible for them to not equal each other.