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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
In triangles APD and BPC; angle ADP = angle BCP Angle ADC = angle BCD = 90° so angle ADP = angle BCP = 60°
Triangles APD and BPC are congruent SAS postulate
What is the error in Ted's proof? (1 point)
He writes the measure of angles ADP and BCP as 60° instead of 45°.
He uses the SAS postulate instead of AAS postulate to prove the triangles congruent.
He writes the measure of angles ADP and BCP as 60° instead of 30°.
He uses the SAS postulate instead of SSS postulate to prove the triangles congruent.
