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Answer:

See Below.

Step-by-step explanation:

We are given that ΔAPB and ΔAQC are equilateral triangles.

And we want to prove that PC = BQ.

Since ΔAPB and ΔAQC are equilateral triangles, this means that:

[tex]PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ[/tex]

Likewise:

[tex]\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ[/tex]

Since they all measure 60°.

Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

[tex]m\angle PAC=m\angle PAB+m\angle BAC[/tex]

Likewise:

[tex]m\angle QAB=m\angle QAC+m\angle BAC[/tex]

Since ∠QAC ≅ ∠PAB:

[tex]m\angle PAC=m\angle QAC+m\angle BAC[/tex]

And by substitution:

[tex]m\angle PAC=m\angle QAB[/tex]

Thus:

[tex]\angle PAC\cong \angle QAB[/tex]

Then by SAS Congruence:

[tex]\Delta PAC\cong \Delta BAQ[/tex]

And by CPCTC:

[tex]PC\cong BQ[/tex]

Answer:

Step-by-step explanation:

To prove PC = BQ, we need to prove triangle APC and ABQ are congruent.

AP = AB as they are part of equilateral triangle APB

AC = AQ as they are part of equilateral triangle AQC

Angle PAC = Angle PAB + Angle BAC = 60 + Angle BAC

Angle BAQ = Angle QAC + Angle BAC = 60 + Angle BAC = Angle PAC

By SAS, triangle APC and ABQ are congruent and therefore

PC = BQ

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