If the diagonals of a quadrilateral bisect the angles, is the quadrilateral always a parallelogram? Explain your answer.

1 answer:

6 0

Yes. If the diagonals bisect the angles, the quadrilateral is always a parallelogram, specifically, a rhombus.

Consider quadrilateral ABCD. If diagonal AC bisects angles A and C, then ΔACB is congruent to ΔACD (ASA). Hence AB=AD and BC=CD (CPCTC).

Likewise, if diagonal BD bisects angles B and D, triangles BDA and BDC are congruent, thus AB=BC and AD=CD. (CPCTC again). Now, we have AB=BC=CD=AD, so the figure is a rhombus, hence a parallelogram.

You might be interested in

Whats the GCF of 10a^3 and 25a?

jeyben [28]

5a, because both are divided by 5 as the greatest common factor, and a to the first power (not a squared or cubed) is the greatest variable power they share :)

5 0

3 years ago

Brandon owes his brother $120. He has a stamp collection worth $1,250 and he has $300 in a bank account. He also has a skateboar

rewona [7]

Answer: brandons assets are the stamp collection, the bank account, and the skate board .

Step-by-step explanation:

4 0

2 years ago

The length and width of a rectangle are measured as 30cm and 24cm, respectively, with an error in measured of

larisa [96]

Answer:

The maximum error in the calculated area of rectangle is 5.4 cm².

Step-by-step explanation:

Given : The length and width of a rectangle are measured as 30 cm and 24 cm, respectively, with an error in measured of at most 0.1 cm in each.

To find : Use differentials to estimate the maximum error in the calculated area of rectangle ?

Solution :

The area of the rectangle is $A=l\times w$

The derivative of the area is equal to the partial derivative of area w.r.t. length times the change in length plus the partial derivative of area w.r.t. width times the change in width.

i.e. $dA=\frac{\partial A}{\partial L}\Delta L+\frac{\partial A}{\partial W}\Delta W$