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.
Please include a diagram if you introduce any new information into your solution.
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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