. (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. Triangle D 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 In triangles APD and BPC; 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) In square ABCD; angle ADC = angle BCD In square ABCD; angle ADP = angle BCP In triangles APD and BPC; angle ADP = angle BCP