In the adjoining figure , APB and AQC are equilateral triangles. Prove that PC = BQ. ( Hint : [tex] \triangle[/tex]APC = [tex] \triangle[/tex]AQB, then PC = BQ)
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Answer:

See Below.

Step-by-step explanation:

Statements:                                                           Reasons:

[tex]\displaystyle 1)\text{ } \Delta APB \text{ and } \Delta AQC \text{ are equilateral triangles}[/tex]      Given

[tex]\displaystyle 2) \text{ } m \angle PAB = 60[/tex]                                                     Definition of equilateral.

[tex]3)\text{ } m \angle QAC = 60[/tex]                                                     Definition of equilateral.

[tex]4)\text{ } m\angle PAB = m\angle QAC[/tex]                                          Substitution

[tex]5)\text{ } m\angle PAC=m\angle PAB+m\angle BAC[/tex]                       Angle Addition

[tex]\displaystyle 6)\text{ } m\angle QAB=m\angle QAC+m\angle BAC[/tex]                       Angle Addition

[tex]7)\text{ } m\angle QAB=m\angle PAB+m\angle BAC[/tex]                       Substitution

[tex]\displaystyle 8)\text{ } m\angle PAC=m\angle QAB[/tex]                                         Substitution

[tex]9)\text{ } PA=BA[/tex]                                                          Definition of equilateral

[tex]10)\text{ } AC=AQ[/tex]                                                        Definition of equilateral

[tex]\displaystyle 11)\text{ } \Delta PAC \cong \Delta BAQ[/tex]                                            Side-Angle-Side Congruence*

[tex]\displaystyle 12)\text{ } PC=BQ[/tex]                                                        CPCTC

* SAS Congruence:

PA = BA

∠PAC = ∠QAB

AC = AQ

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