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

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

StatementsJustifications
In triangles APD and BPC; DP = PCSides of equilateral triangle DPC are equal
Sides of square ABCD are equal
In triangles APD and BPC; angle ADP = angle BCPAngle ADC = angle BCD = 90° and angle ADP = angle BCP = 90° − 60° = 30°
Triangles APD and BPC are congruentSAS postulate
Which of the following completes Jake's proof? (1 point)

In triangles APD and BPC; AD = BC

In triangles APD and BPC; AP = PB

In triangles APB and DPC; AD = BC

In triangles APB and DPC; AP = PB