Question

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

Answer 1

Answer:

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\