The figure below shows a square abcd and an equilateral triangle dpc: nick makes the chart shown below to prove that triangle apd is congruent to triangle bpc: statements justifications in triangles apd and bpc; ap = pb p is the vertex of the equilateral triangle 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 sas postulate what is the error in nick's proof? he uses the sas postulate instead of aas postulate to prove the triangles congruent. he assumes that ap = pb because p is the vertex of the equilateral triangle. he uses the sas postulate instead of sss postulate to prove the triangles congruent. he writes the measure of angles adp and bcp as 30° instead of 45°.

The only incorrect statement in the Nick's proof is : He assumes that AP = PB because P is the vertex of the equilateral triangle.

Step-by-step explanation:

To show : ΔAPD ≅ ΔBPC

Given : ΔDPC is an equilateral triangle.

Then, DP = CP (Same side of the equilateral triangle)

 ∠ ADP = ∠ ADC - ∠ PDC = 90° - 60° = 30°

Similarly,

∠ BCP = ∠ BCD - ∠ PCD = 90° - 60° = 30°

∴ ∠ ADP = ∠ BCP

and AD = BC (Same side of square)

∴ Δ APD ≅ Δ BPC ( SAS proved)

So, the only incorrect statement in the Nick's proof is : He assumes that AP = PB because P is the vertex of the equilateral triangle.

Because, if P is vertex of equilateral triangle then it doesn't implies that AP is equal to PB

(B)He assumes that AP = PB because P is the vertex of the equilateral triangle.  

Step-by-step explanation:

From the ΔAPD and ΔBPC, we have

AD=BC (sides of the square)

∠ADP=∠BCP( Each 30°)

PD=PC (sides of equilateral triangle)

Thus, by SAS rule, ΔAPD ≅ ΔBPC,

Since, Nick has taken AP=PB because P is the vertex of equilateral triangle instead of PD=PC which does not help to prove that ΔAPD ≅ ΔBPC because he can't take such assumptions which are not valid.

Thus, he should take PD=PC which are the sides of equilateral triangle and are equal. Therefore, Option B is the error in Nick's solution.

the graph g(x) is f(x) translated 7 units in the negative y direction and translated 2 units in the negative x direction