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kotykmax [81]
3 years ago
3

. (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 squ

are. 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
Mathematics
2 answers:
MAVERICK [17]3 years ago
7 0

Answer:  \angle ADP = \angle BCP

Step-by-step 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



-Dominant- [34]3 years ago
6 0

Answer:

Δ APD≅Δ BPC

Step-by-step explanation:

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