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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.


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 class=

Sagot :

Answer:

Which of the following completes Ted's proof?

In square ABCD; angle ADC = angle BCD

In square ABCD; angle ADP = angle BCP

In triangles APD and BPC; angle ADC = angle BCD

In triangles APD and BPC; angle ADP = angle BCP\

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