The following is an incomplete flowchart proving that the opposite angles of parallelogram JKLM are congruent:
2 answers:
Answer:
(B) corresponding angles theorem
Step-by-step explanation:
Given: A parallelogram JKLM is shown where segment JM is parallel to segment KL and segment JK is parallel to segment ML.
To prove: Opposite angles of parallelogram are equal.
Construction: Extend segment JM beyond point M and draw point P, Extend segment JK beyond point J and draw point Q.
Proof:
It is given that A parallelogram JKLM is shown where segment JM is parallel to segment KL and segment JK is parallel to segment ML.
Extend segment JM beyond point M and draw point P--------By construction
and Extend segment JK beyond point J and draw point Q----By construction
thus, ∠MLK≅∠PML and ∠JML≅∠QJM (Alternate interior angles theorem) (1)
Then, ∠PML≅∠KJM and ∠QJM≅∠LKJ (corresponding angles theorem) (2)
Using equation (1) and (2) and using the transitive property of equality, we have
∠MLK≅∠KJM and ∠JML≅∠LKJ
therefore, opposite angles of the given parallelogram JKLM are congruent.
Hence proved.
You might be interested in
To write the system we need the slope of each line and at least one point on the line. The two lines to consider will be the lines connecting the location of each plane to the airport they are flying to. It is also worth noting that the coordinates of the airport represent the point of intersection of the two lines and thus the solution to the system.
1. slope of the line connecting airplane one and the airport: m = 2 you can see this clearly if you graph the two points. From airplane 1 location we rise 8 units and move to the right 4 units to get to the airport. Slope is defined as rise over run: so 8 divided by 4 = 2(the slope) Now substitute the slope and the point (2,4) into point-slope form of a line:
y - 4 = 2(x -4) the standard form of this equation is 2x - y = 0
2. slope of the line connecting airplane 2 and the airport: m = -
To find this slope, simply observe the vertical change of down 3 and a horizontal shift of right 9 from the airport to airplane 2. Now substitute this slope and and the point (15,9) into point-slope form of a line:
y - 9 =
(x - 15) the standard form of this equation is:
x + 3y = 42
Let's write the system:
2x - y = 0
x + 3y = 42
Multiply the first equation by 3 to get the new system
6x - 3y = 0
x + 3y = 42 add these two equations to get an equation in terms of x
7x = 42 thus x = 6 and substituting this value into 2x - y = 0 we see y = 12
In other words, we have proven that the location of the airport is in fact the solution to our system.
PS: You just have to do a little algebra to get from point-slope form of the two equations to standard form. I did not show this process, but if you need it just let me know... thanks
1. slope of the line connecting airplane one and the airport: m = 2 you can see this clearly if you graph the two points. From airplane 1 location we rise 8 units and move to the right 4 units to get to the airport. Slope is defined as rise over run: so 8 divided by 4 = 2(the slope) Now substitute the slope and the point (2,4) into point-slope form of a line:
y - 4 = 2(x -4) the standard form of this equation is 2x - y = 0
2. slope of the line connecting airplane 2 and the airport: m = -
y - 9 =
x + 3y = 42
Let's write the system:
2x - y = 0
x + 3y = 42
Multiply the first equation by 3 to get the new system
6x - 3y = 0
x + 3y = 42 add these two equations to get an equation in terms of x
7x = 42 thus x = 6 and substituting this value into 2x - y = 0 we see y = 12
In other words, we have proven that the location of the airport is in fact the solution to our system.
PS: You just have to do a little algebra to get from point-slope form of the two equations to standard form. I did not show this process, but if you need it just let me know... thanks