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.
I am not good at presenting proofs properly but here is my logic.
Let X be the centr of the circle.
Equal sectors cut of equal segments so the perpendicular heights from the centre of each segment chord to the circumference is going to be equal.
AX=BX=CX=DX=EX=FX all equal radii
\(\triangle ABX \cong \triangle CDX \cong\triangle EFX\)
and all these triangles are isoscles with AB, CD, and EF respectively being the base sides.
The perpendicular bisector of the base sides will pass throught the centre of the circle. They are all altitudes off the base sides.
Since the altitudes are all the same, the circle that passes through these bisector points is concentric with the original circle.
Hence the centre of the incircle is also the centre of the original circle.
This definitely needs to be tidied up but do you understand what I am saying?
Draw perpendicular bisectors to AB, FE and CD....since these are equal chords in the larger circle, by Euclid, equal chords in a circle are equidistant from the center of the larger circle.......
Let O be this center and let the points where the perpendicular bisectors intersect AB, FE and CD be labeled as X, Y and Z, respectively......and....as before, OX = OY = OZ and a smaller circle can be constructed with these as radii............but.....this must mean that O is also the center of this circle, as well....and... the largest circle that will fit inside the triangle will touch all three sides, so, by definition, O is also the incenter of PQR