As, AC ≠ BD. Moreover, 2 diagonals space perpendicular to each other in a rhombus. Thus, AC ⊥ BD.

You are watching: Diagonals ac and bd form right angles at point m in parallelogram abcd. prove abcd is a rhombus.

Hence, we acquire that, alternative A is correct.



Answer from: michellemonroe012305

1. AC ⊥BD ; therefore, ABCD is a rhombus.

Step-by-step explanation:

We space given,

A parallelogram ABCD with vertices A(1,2), B(0,9), C(7,8) and also D(8,1).

Using street Formula, which offers the distance between two points (

*
and (
*
as
*

So, the distance in between two clues are given by,

AB=

*
= 7.07

BC=

*
= 7.07

CD=

*
= 7.07

DA=

*
= 7.07

Also,

AC=

*
= 8.49

BD=

*
= 11.3

Since, we have,

AB = BC = CD = DA.

So, ABCD is a rhombus yet not a rectangle.

As, AC ≠ BD. Moreover, two diagonals room perpendicular come each various other in a rhombus. Thus, AC ⊥ BD.

Hence, we obtain that, choice A is correct.



Answer from: amulets7017

B) ABCD is a rhombus together all the sides space congruent (equal)



Answer from: ubaldo7410

B

Step-by-step explanation:



Answer from: kordejah348

C. ABCD is a rhombus, yet not necessarily a square.

Step-by-step explanation:

A parallel with surrounding sides congruent is a rhombus, which additionally happens to have perpendicular diagonals (that bisect each other). If the diagonals are various length, then the parallelogram is not a rectangle, therefore the rhombus is not a square.



Answer from: savannahsharp5981

Option C is correct.

Explanation:

Rhombus states that a parallel with four equal sides and also sometimes one with no best angle.

Given: The coordinate of the vertices of quadrilateral ABCD space A(−6, 3) , B(−1, 5) , C(3, 1) , and D(−2, −2) .

The condition for the segment

*
,
*
to it is in parallel to
*
,
*
is equivalent slopes;

*
or

*
....<1>

So, we have to examine that and

First check

A(−6, 3) , B(−1, 5) , C(3, 1) , and also D(−2, −2)

substitute in <1>,

*

*

-10 ≠ -15

Similarly,

check

A(−6, 3) , D(−2, −2) , B(−1, 5) and also C(3, 1)

Substitute in <1>, we have

*

*

-20 ≠ -16.

Both bag of sides are not parallel,

therefore, quadrilateral ABCD is not a rhombus because there are no pairs of parallel sides.


Answer from: rocksquad7917

Option D

Step-by-step explanation:

In any Rhombus the diagonals bisect the angles. The diagonals are perpendicular bisectors of every other.

So,

5x-18+x+90=180 ( angles of a triangle add to 180 degrees)

Simplifying prefer terms:

6x+72=180

Subtracting 72 both sides :

6x= 108

Dividing through 6 both sides:

x=18.

Option D is correct.


Answer from: joyelewis58

Solution:

Vertices of parallelogram ABCD is provided as A(1,2) , B(0,9) , C(7,8) , and D(8,1) .

We will usage the following analytical geometry formulas here

1. Distance between two points

*

2. Slope of a line

*

3. When two lines room perpendicular, product of their slopes is same to -1.

Since ABCD is a parallelogram

1. The contrary sides space equal and also parallel.

2. Diagonals bisect each other.

3. Opposite angles space equal.

Now, comes to problem

*

As, you can see that, AB=BC=CD=DA=√ 50

But , steep of ab × steep of BC =slope the CB × slope of DC=slope that CD × slope of DA

*
=1" />

which is no equal come -1. It method lines which room sides the parallelogram space not perpendicular.

As, every side of parallelogram ABCD space equal, so the is a rhombus.

As, diagonal line of rhombus bisect each various other at appropriate angles.

*

Shows the diagonals are perpendicular bisector of every other.

Option (1) : AC⊥BD; therefore, ABCD is a rhombus.

*

Answer from: HoodieHeem

Solution:

Vertices of parallel ABCD is offered as A(1,2) , B(0,9) , C(7,8) , and D(8,1) .

We will use the following analytical geometry formulas here

1. Distance between two points

*

2. Steep of a line

*

3. Once two lines space perpendicular, product of your slopes is same to -1.

Since ABCD is a parallelogram

1. The contrary sides space equal and also parallel.

2. Diagonals bisect every other.

3. The opposite angles space equal.

Now, coming to problem

*

As, you deserve to see that, AB=BC=CD=DA=√ 50

But , steep of abdominal × steep of BC =slope of CB × slope of DC=slope the CD × steep of DA

*
=1" />

which is no equal to -1. It way lines which room sides that parallelogram room not perpendicular.

As, every side of parallel ABCD space equal, so the is a rhombus.

As, diagonal of rhombus bisect each various other at appropriate angles.

*

Shows that diagonals are perpendicular bisector of every other.

Option (1) : AC⊥BD; therefore, ABCD is a rhombus.

*

Answer from: jj1077348

Step-by-step explanation:

Refer the attached number :

In Δ ABD ,

∠B=∠D = 66°

By building : opposite sides of equal angles are equal

Thus AB=AC

In Δ CBD ,

∠B=∠D = 66°

By building : opposite political parties of equal angles space equal

Thus CB=CD

Thus all 4 sides of quadrilateral ABCD room equal

And diagonal line BD bisects the angle

So, it is a rhombus

Thus choice c is exactly .

c. Parallel ABCD is a rhombus, due to the fact that the diagonal bisects two angles.

*

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