(05.03 MC) The figure below shows a square ABCD and an equilateral triangle DPC:

ABCD is a square. P is a point inside the square. Straight lines join points A and P, B and P, D and P, and C and P.

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

| Statements | Justifications |
|---------------------------------------------|--------------------------------------------------------|
| In triangles APD and BPC: | |
| DP = PC | Sides of equilateral triangle DPC are equal |
| AD = BC | 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)

A. In square ABCD; angle ADC = angle BCD

B. In square ABCD; angle ADP = angle BCP

C. In triangles APD and BPC; angle ADP = angle BCP