Question

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

~Thanks in advance ! <3​

Answer 1

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

Answer 2

See Below.

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`