Question

The figure below shows a square ABCD and an equilateral triangle DPC:

Ted makes the chart shown below to prove that triangle APD is congruent to triangle BPC:

Which of the following completes Ted's proof?
In square ABCD; angle ADC = angle BCD
In square ABCD; angle ADP = angle BCP
In triangles APD and BPC; angle ADC = angle BCD
In triangles APD and BPC; angle ADP = angle BCP