(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