Question

Question : In the given figure , ∆ APB and ∆ AQC are equilateral triangles. Prove that PC = BQ.

Answer 1

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:

$PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ$

Likewise:

$\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ$

Since they all measure 60°.

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

$m\angle PAC=m\angle PAB+m\angle BAC$

Likewise:

$m\angle QAB=m\angle QAC+m\angle BAC$

Since ∠QAC ≅ ∠PAB:

$m\angle PAC=m\angle QAC+m\angle BAC$

And by substitution:

$m\angle PAC=m\angle QAB$

Thus:

$\angle PAC\cong \angle QAB$

Then by SAS Congruence:

$\Delta PAC\cong \Delta BAQ$

And by CPCTC:

$PC\cong BQ$

Answer 2

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