The figure below shows a square ABCD and an equilateral triangle DPC: D
Ted makes the chart shown below to prove that triangle APD is congruent to triangle BPC:
Statements
In triangles APD and BPC; DP = PC
In triangles APD and BPC; AD = BC
Justifications
Sides of equilateral triangle DPC are equal
Sides of square ABCD are equal
Angle ADC = angle BCD = 90° so angle ADP = angle BCP = 30°
Triangles APD and BPC are congruent SAS postulate
Which of the following completes Ted's proof? (1 point)
O In square ABCD; angle ADC= angle BCD
O In square ABCD; angle ADP= angle BCP
O In triangles APD and BPC; angle ADC
angle BCD
O In triangles APD and BPC; angle ADP = angle BCP

whose power in the senate is equal to the power of the speaker of the house?

step-by-step explanation:

when rotating a figure 180 degrees, imagine that you are able to take the figure and turn it clockwise or counterclockwise around a center point. to rotate a figure 180 degrees, you will need to apply the rule (x, y) → (-x, -y). start by using a coordinate grid with coordinates for each vertex of the figure. the center point of the coordinate grid is located at (0, 0), which is what you will rotate the figure around. write down the original coordinates of the shape you are going to rotate. then, apply the rule. for example, coordinate (1, 2) becomes (-1, -2). another example would be, coordinate (-4, -2) becoming (4, 2). after you change each original coordinate to the rotated coordinate, you will draw your new figure.

step-by-step explanation:

the answer is a beach is the best place to be