Question

Step 2: Construct regular polygons inscribed in a circle.

Answer 1

Firstly, we can observe that line segment PQ is the radius of Circle Q, as well as the fact that line segment PQ is the radius of Circle P. With this observation we can confirm that Circles P and Q have the same radius, and that they are equal. Next, we are also able to observe that line segment RQ is a radius of Circle Q, so we can prove that line segments PQ and RQ are equal due to them both being radii of Circle Q. Similarly, line segment PR is also a radius of Circle P, so we can prove that line segments PQ and PR are equal due to them both being radii of Circle P. Finally, due to the transative property we can prove that line segments PR and RQ are equal, so that means that all sides of Triangle PQR, (PQ, RQ, and PR) are equal, which means that Triangle PQR is and equilateral Triangle.

Answer 2

Several ways to prove but let's use one only

Here

• PQ is the radius of circle Q

Also

• PQ is the radius of circle P

Hence P and Q circles have same radius

Now

• PQ=RQ

As both are radius of circle Q

And

• PQ=PR

As both are radius of circle P.

Hence as per transitive property

• PR=QR

Now

• PQ=QR=PR

Hence

∆PQR is equilateral