Question

. (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

Answer 1

ADP = BCP

Explanation:

Answer 2

Answer:
`\angle ADP = \angle BCP`

Explanation:

Here, ABCD is a square, and P is a point inside the square.

Where, Δ DPC is a equilateral triangle.

Therefore, DP=PC

Thus In triangles APD and BPC,

DP= CP

`\angle ADP= \angle BCP` ( because both are of 30°)

AD=CB ( sides of square)

Therefore by SAS postulate,

Δ APD≅Δ BPC