Question

The figure below shows a square ABCD and an equilateral triangle DPC:

Answer 1

Jim 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° and angle ADP = angle BCP = 90° - 60° = 30°
Triangles APD and BPC are congruent SSS postulate

Answer 2

What is the error in Jim's proof?
He writes DP = PC instead of DP = PB.

He writes AD = BC instead of AD = PC.

He assumes the measure of angle ADP and angle BCP as 30° instead of 45°.

He assumes that the triangles are congruent by the SSS postulate instead of SAS postulate.

Answer 3

@CO_oLBoY @mesanyam10

Answer 4

need help please!

Answer 5

the only errors i see are that he wrote DP = PC instead of DP = CP and he should use SAS instead of SSS

Answer 6

angles ADP and BCP are 30 since the angles PCD and PDC are 60 (since it's an equilateral triangle)

Answer 7

so the answer is A? @mtbender74

Answer 8

nope

Answer 9

which one then I am confused? @CO_oLBoY

Answer 10

coz A states that 'He writes DP = PC instead of DP = PB.' which isnt correct as infact DP=PC

Answer 11

So what's the answer which choice? @CO_oLBoY

Answer 12

Use ΔABC and ΔEDF shown below to answer the question that follows:

Answer 13

If AB over ED equals BC over DF and ∠B ≅ ∠D, prove that ΔABC is similar to ΔEDF.

Answer 14

Need help with one more by the way! :)

Answer 15

i think ur first question choice is D and lemme look ur 2nd question

Answer 16

ok @CO_oLBoY

Answer 17

∠B ≅ ∠D given
as
AB over ED equals BC over DF
so their respective angles are equal
so, ∠A ≅ ∠E
and ∠C ≅ ∠F
if all three angles of ΔABC and ΔEDF are equal then
ΔABC is similar to ΔEDF

Answer 18

thanks @CO_oLBoY