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.

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.

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

B

Step-by-step explanation:

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.

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.

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.

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.

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.

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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