Triangle Bisectors

Find the center of the circle.....call this  " O"

From O.....draw  bisector OG  to chord AB   and bisector OH  to chord EF 

But, by Euclid,  a radial line bisecting a chord does so at right angles

And since AB  = EF = CD.....then equal chords are equal distances from the center of a circle

Now....connect  O  and  P

So.....OG  = OH,  OP  = OP   and angles OGP  and OHP  are right

So, by HL,  have two  cogruent  right triangles formed, ΔOGP  and ΔOHP

But  angle GPO   =  angle HPO....so  angle P  is bisected

By similar reasoning, we can draw bisector  OI  to chord  CD  and  also connect  QO

This will form  two congruent right triangles - Δ OGQ and ΔOIQ

And angle GQO  =  angle IQO, so Q is bisected

And since PO  and QO are angle bisectors of triangle  QPR  that  meet  at O....then ,by definition, the center of the circle is the incenter of triangle PQR