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