A circle is drawn that intersects all three sides of triangle PQR as shown below. Prove that if AB = CD = EF, then the center of the circle is the incenter of triangle PQR.
Take O as the center of the circle
Draw OG, OH, OI perpendicular to AB, CD and EF respectively
And, by Euclid.....equal chords are equal distances from the center of a circle
So GO = HO and OQ is common
So triangles QOG and QOH are congruent right triangles by HL
So < GQO = < HQO
By similar reasoning, triangles ROH and ROI are congruent right triangles by HL
So < HRO = < IRO
And once more, triangles POI and POG are congruent right triangles by HL
So < IPO = < GPO
Therefore......since QO, RO and PO are angle bisectors of triangle PQR that meet at O, then O is the triangle incenter.....but O is also the center of the circle