Its an indirect proof, so 3 steps :-
1) you start with the opposite of wat u need to prove
2) arrive at a contradiction
3) concludeReport · 29/6/2015261
since you wanto prove 'diagonals of a parallelogram bisect each other', you start wid the opposite of above statement, like below :- step1 : Since we want to prove 'diagonals of a parallelogram bisect each other', lets start by assuming the opposite, that the diagonals of parallelogram dont bisect each other.Report · 29/6/2015261

Since, we assumed that the diagonals dont bisect each other,
OC≠OA
OD≠OBReport · 29/6/2015261
Since, OC≠OA, △OAD is not congruent to △OCBReport · 29/6/2015261

∠AOD≅∠BOC as they are vertical angles,
∠OAD≅∠OCB they are alternate interior angles
AD≅BC, by definition of parallelogram
so, by AAS, △OAD is congruent to △OCBReport · 29/6/2015261
But, thats a contradiction as we have previously established that those triangles are congruentReport · 29/6/2015261
step3 :
since we arrived at a contradiction, our assumption is wrong. so, the opposite of our assumption must be correct. so diagonals of parallelogram bisect each other.

4 0

2 years ago

A rectangular lot is 70 yards wide and 115 yards long.

forsale [732]

Answer: 165 in length 20 in width

Step-by-step explanation: :)

5 0

1 year ago

Bob can't believe how bad London's traffic is today. In the past 10 minutes, his cab has barely moved 500 meters! He wonders if

PolarNik [594]

I’ll answer your question just help me with mine pls7:7

8 0

2 years ago

Read 2 more answers

Need help with homework

motikmotik

We have two points describing the diameter of a circumference, these are:

$\begin{gathered} A=(-12,-4) \\ B=(-4,-10) \end{gathered}$

Recall that the equation for the standard form of a circle is:

$(x-h)^2+(y-k)^2=r^2$

Where (h,k) is the coordinate of the center of the circle, to find this coordinate, we find the midpoint of the diameter, that is, the midpoint between points A and B.

For this we use the following equation:

$M=(\frac{x_1+x_2_{}_{}}{2},\frac{y_1+y_2}{2})$

Now, we replace and solve:

$\begin{gathered} M=(\frac{-12+(-4)}{2},\frac{-4+(-10)}{2} \\ M=(\frac{-12-4}{2},\frac{-4-10}{2}) \\ M=(\frac{-16}{2},\frac{-14}{2}) \\ M=(-8,-7) \end{gathered}$

The center of the circle is (-8,-7), so:

$\begin{gathered} h=-8 \\ k=-7 \end{gathered}$

On the other hand, we must find the radius of the circle, remember that the radius of a circle goes from the center of the circumference to a point on its arc, for this we use the following equation:

$r^2=\Delta x^2+\Delta y^2$

In this case, we will solve the delta with the center coordinate and the B coordinate.

$\begin{gathered} r^2=((-4)-(-8))^2+((-10)-(-7)) \\ r^2=(-4+8)^2+(-10+7)^2 \\ r^2=4^2+(-3)^2 \\ r^2=16+9 \\ r^2=25 \\ r=5 \end{gathered}$

Therefore, the equation for the standard form of a circle is:

$\begin{gathered} (x-(-8))^2+(y-(-7))^2=25 \\ (x+8)^2+(y+7)^2=25 \end{gathered}$

In conclusion, the equation is the following:

$(x+8)^2+(y+7)^2=25$

4 0

2 months ago